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Class 11 NCERT Solutions

Chapter 9: Straight Lines

Master the various forms of equations, the shifting of origins, and the logic of linear gradients with our step-by-step logic.

Exercise 9.1

Q1. Draw a quadrilateral in the Cartesian plane, whose vertices are (- 4, 5), (0, 7), (5, – 5) and (- 4, – 2). Also, find its area.

Sol. Recall that the area of quadrilateral ABCD
$=$ Area of triangle $\mathrm{ABD}+$ Area of triangle BCD

[Using the formula for area of triangle
$
=\frac{1}{2}\left|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right|
$
for both triangles ABD and BCD ]
Area of quadrilateral ABCD
$
\begin{aligned}
= & \frac{1}{2}|-4(7+2)+0(-2-5)-4(5-7)| \\
& +\frac{1}{2}|0(-5+2)+5(-2-7)-4(7+5)| \\
= & \frac{1}{2}|-36+8|+\frac{1}{2}|-45-48| \\
= & \frac{1}{2}(28)+\frac{1}{2}(93)=14+46.5 \\
= & 60.5 \text { square units. }
\end{aligned}
$

Q2. The base of an equilateral triangle with side $2 a$ lies along the $y$-axis such that the mid-point of the base is at the origin. Find vertices of the triangle.

Sol. Suppose the base AB of equilateral triangle ABC (side $2a$) lies along the $y$-axis, with the origin O as the midpoint of AB. Therefore, $\mathrm{OA}=\mathrm{OB}=a$ and co-ordinates of A are ( $0, a$ ) and those of B are ( $0,-a$ ). ( $\because$ on $y$-axis $x=0$ ) Since side AB is $y$-axis (given) and O , the mid-point of AB , therefore OC is the right bisector of $\mathrm{AB}(\because \triangle \mathrm{ABC}$ is an equilateral) and is $x$-axis.

Now
$
\mathrm{OC}=\sqrt{\mathrm{AC}^2-\mathrm{OA}^2}
$
(By Pythagoras Theorem)
$
=\sqrt{(2 a)^2-a^2}=\sqrt{3 a^2}=\sqrt{3} a
$

Similarly $\mathrm{OC}^{\prime}=\sqrt{3} a$.
∴ The co-ordinates of the third vertex C are $(\sqrt{3} a, 0)$ and those of $C^{\prime}$ are ( $-\sqrt{3} a, 0$ ).

Q3. Find the distance between $\mathbf{P}\left(x_1, y_1\right)$ and $\mathbf{Q}\left(x_2, y_2\right)$ when: (i) $\mathbf{P Q}$ is parallel to the $\boldsymbol{y}$-axis, (ii) $\mathbf{P Q}$ is parallel to the $\boldsymbol{x}$-axis.

Sol. (i) Since PQ is parallel to the $y$-axis, we have $x_1=x_2$.
$
\left.\begin{array}{rl}
\therefore \mathrm{PQ} & =\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2} \\
& =\sqrt{\left(y_2-y_1\right)^2}=\left|y_2-y_1\right| \\
{\left[\because \sqrt{x^2}=|x|\right]}
\end{array}\right\}
$
(ii) Since PQ is parallel to the $x$-axis, we have $y_1=y_2$.
$
\begin{aligned}
\therefore \mathrm{PQ} & =\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2} \\
& =\sqrt{\left(x_2-x_1\right)^2}=\left|x_2-x_1\right|
\end{aligned}
$

Q4. Find a point on the $x$-axis, which is equidistant from the points ( 7,6 ) and ( 3,4 ).

Sol. Let $\mathrm{P}(x, 0)$ be a point on the $x$-axis (where $y=0$) equidistant from $\mathrm{A}(7,6)$ and $\mathrm{B}(3,4)$.

Then
$
\mathrm{PA}=\mathrm{PB}
$
$
\Rightarrow \quad \sqrt{(x-7)^2+(0-6)^2}=\sqrt{(x-3)^2+(0-4)^2}
$

Squaring, $\quad x^2-14 x+49+36=x^2-6 x+9+16$
$
\Rightarrow \quad-14 x+85=-6 x+25
$
$
\Rightarrow \quad-8 x=-60 \quad \therefore \quad x=\frac{60}{8}=\frac{15}{2}
$
∴ The required point on $x$-axis is $\left(\frac{15}{2}, 0\right)$.

Q5. Find the slope of a line, which passes through the origin, and the mid-point of the line segment joining the points $\mathbf{P}(\mathbf{0},-\mathbf{4})$ and $\mathbf{B}(\mathbf{8}, \mathbf{0})$.

Sol. Let M be the midpoint of PB. By the midpoint formula:
$
\mathrm{M}=\left(\frac{0+8}{2}, \frac{-4+0}{2}\right)=(4,-2)
$

Since the origin O has coordinates $(0,0)$,
$
\text { ∴ Slope of OM }=\frac{y_2-y_1}{x_2-x_1}=\frac{-2-0}{4-0}=\frac{-2}{4}=-\frac{1}{2} \text {. }
$

Q6. Without using the Pythagoras theorem, show that the points $(4,4),(3,5)$ and $(-1,-1)$ are the vertices of a right-angled triangle.

Sol. Label the given points as $\mathrm{A}(4,4)$, $\mathrm{B}(3,5)$ and $\mathrm{C}(-1,-1)$.
$
\begin{aligned}
& m_1=\text { slope of } \mathrm{AB}=\frac{y_2-y_1}{x_2-x_1}=\frac{5-4}{3-4}=\frac{1}{-1}=-1 \\
& m_2=\text { slope of } \mathrm{BC}=\frac{-1-5}{-1-3}=\frac{-}{-4}=\frac{3}{2} \\
& m_3=\text { slope of } \mathrm{AC}=\frac{-1-4}{-1-4}=\frac{-5}{-5}=1
\end{aligned}
$

Notice that $m_1 \cdot m_3=(-1)(1)=-1$.
∴ Lines AB and AC are perpendicular to each other.
∴ Triangle ABC is a right-angled triangle with angle $\mathrm{A}=90^{\circ}$.
∴ The given points are the vertices of a right-angled triangle.

Q7. Find the slope of the line, which makes an angle of $30^{\circ}$ with the positive direction of $y$-axis measured anticlockwise.

Sol. The line is inclined at $30^{\circ}$ to the positive $y$-axis (measured anticlockwise from OY).
∴ The line makes an angle $90^{\circ}+30^{\circ}=120^{\circ}$ with the positive direction of $x$-axis.
∴ Slope of the line
$
\begin{aligned}
& =\tan 120^{\circ} \\
& =\tan \left(180^{\circ}-60^{\circ}\right) \\
& =-\tan 60^{\circ}=-\sqrt{3} .
\end{aligned}
$

Q8. Find the value of $x$ for which the points ( $x,-1$ ), ( 2,1 ) and ( 4,5 ) are collinear.

Sol. Since $\mathrm{A}(x,-1)$, $\mathrm{B}(2,1)$ and $\mathrm{C}(4,5)$ are collinear,
the slope of AB must equal the slope of BC:
$
\begin{aligned}
& \Rightarrow & \frac{1-(-1)}{2-x} & =\frac{5-1}{4-2} & & \\
\Rightarrow & \frac{2}{2-x} & =2 & & \Rightarrow & 2=4-2 x \\
& \Rightarrow & 2 x & =2 & & \therefore x=1 .
\end{aligned}
$

Q9. Without using distance formula, show that points $(-2,-1),(4,0),(3,3)$ and $(-3,2)$ are the vertices of a parallelogram.

Sol. Let the four points be $\mathrm{A}(-2,-1)$, $\mathrm{B}(4,0)$, $\mathrm{C}(3,3)$ and $\mathrm{D}(-3,2)$.

Slope of $\mathrm{AB}=\frac{y_2-y_1}{x_2-x_1}=\frac{0-(-1)}{4-(-2)}=1$.
Slope of $\mathrm{DC}=\frac{3-2}{3-(-3)}=1$.
⇒ Slope of $\mathrm{AB}=$ Slope of DC .
$\Rightarrow \quad \mathrm{AB} \| \mathrm{DC}$

Slope of $\mathrm{BC}=\frac{3-0}{3-4}=\frac{3}{-1}=-3$
Slope of $\mathrm{AD}=\frac{2-(-1)}{-3-(-2)}=\frac{2+1}{-3+2}=\frac{3}{-1}=-3$
⇒ Slope of $\mathrm{BC}=$ Slope of AD
$\Rightarrow \quad \mathrm{BC} \| \mathrm{AD}$

Since both pairs of opposite sides are parallel, ABCD is a parallelogram.

Q10. Find the angle between the $\boldsymbol{x}$-axis and the line joining the points (3, -1) and (4, -2).

Sol. The slope of line AB through $\mathrm{A}(3,-1)$ and $\mathrm{B}(4,-2)$ is
$
\frac{-2-(-1)}{4-3}=\frac{-2+1}{1}=-1 .
$

If AB makes angle $\theta$ with the positive $x$-axis, then its slope equals $\tan \theta$.
$\therefore \tan \theta=-1=-\tan 45^{\circ}=\tan \left(180^{\circ}-45^{\circ}\right)=\tan 135^{\circ}$
$\Rightarrow \quad \theta=135^{\circ}$.

Q11. The slope of a line is double of the slope of another line. If tangent of the angle between them is $\frac{\mathbf{1}}{\mathbf{3}}$, find the slopes of the lines.

Sol. Let the two slopes be $m$ and $2m$. If $\theta$ is the angle between the lines, then
$
\begin{gathered}
\tan \theta=\frac{1}{3} \\
\text { Using } \tan \theta=\left|\frac{m_1-m_2}{1+m_1 m_2}\right| \\
\Rightarrow \quad\left|\frac{2 m-m}{1+2 m \cdot m}\right|=\frac{1}{3} \Rightarrow\left|\frac{m}{1+2 m^2}\right|=\frac{1}{3} \\
\Rightarrow \quad \frac{m}{1+2 m^2}= \pm \frac{1}{3} \quad(\therefore|x|=a(a \geq 0) \Rightarrow x \pm a)
\end{gathered}
$

When $\frac{m}{1+2 m^2}=\frac{1}{3}$, we have
$
\begin{array}{lclr}
& 3 m=1+2 m^2 . & & \\
\Rightarrow & 2 m^2-3 m+1=0 & \Rightarrow & 2 m^2-2 m-m+1=0 \\
\Rightarrow & 2 m(m-1)-(m-1)=0 & \Rightarrow & (m-1)(2 m-1)=0 \\
\Rightarrow & m=1, \frac{1}{2} & &
\end{array}
$

When $\frac{m}{1+2 m^2}=-\frac{1}{3}$, we have
$
\begin{array}{rlrl}
-3 m & =1+2 m^2 \\
\Rightarrow & 2 m^2+3 m+1 & =0 \\
\Rightarrow & m & =-1,-\frac{1}{2}
\end{array} \quad \Rightarrow \quad(m+1)(2 m+1)=0
$

Combining all cases, $m=1, \frac{1}{2},-1,-\frac{1}{2}$, giving slope pairs: $(1,2)$, $(\frac{1}{2},1)$, $(-1,-2)$, or $(-\frac{1}{2},-1)$.

Q12. A line passes through $\left(x_1, y_1\right)$ and $(h, k)$. If slope of the line is $m$, show that $k-y_1=m\left(h-x_1\right)$.

Sol. The slope of the line through $\left(x_1, y_1\right)$ and $(h, k)$ is
$
\frac{k-y_1}{h-x_1}=m
$
(given)
Cross-multiplying gives $k-y_1=m\left(h-x_1\right)$.

Q13. If three points $(h, 0),(a, b)$ and $(0, k)$ lie on a line, show that $\quad \frac{\boldsymbol{a}}{\boldsymbol{h}}+\frac{\boldsymbol{b}}{\boldsymbol{k}}=\mathbf{1}$.

Sol. Since $\mathrm{P}(h,0)$, $\mathrm{Q}(a,b)$ and $\mathrm{R}(0,k)$ are collinear,
the slope of PQ must equal the slope of QR:
$
\begin{array}{rlrl}
\Rightarrow & \frac{b-0}{a-h} & =\frac{k-b}{0-a} & \Rightarrow \\
\Rightarrow & -a b & =(a-h)(k-b) & \Rightarrow \\
\Rightarrow & 0 & =a k=a b=a k-a b-h k+h b \\
& =a k+h b & \Rightarrow & h k=a k+h b
\end{array}
$

Dividing both sides by $hk$,
$
1=\frac{a}{h}+\frac{b}{k} \quad \text { or } \quad \frac{a}{h}+\frac{b}{k}=1 .
$

Q14. Consider the following population and year graph (see figure), find the slope of the line AB and using it, find what will be the population in the year 2010?

Sol. The slope of line AB is $\frac{97-92}{1995-1985}=\frac{5}{10}=\frac{1}{2}$.
Let $p$ (crores) be the population in 2010; then point $\mathrm{C}(2010, p)$ lies on line AB.

∴ Slope of $\mathrm{AB}=$ Slope of BC .
$
\begin{array}{rlrlrl}
& \Rightarrow & \frac{1}{2} & =\frac{p-97}{2010-1995} & & \Rightarrow \\
\Rightarrow & & \frac{15}{2} & =p-97 \\
& \Rightarrow & & \Rightarrow & 15 \\
& & p & =104.5 & &
\end{array}
$

Therefore, the estimated population in 2010 is 104.5 crores.

Exercise 9.2

Q1. Write the equations for the $\boldsymbol{x}$-and $\boldsymbol{y}$-axis.

Sol. For any point $\mathrm{P}(x,y)$ on the $x$-axis, the ordinate is always zero, so $y=0$.
Therefore, the equation of the $x$-axis is $y=0$.
For any point on the $y$-axis, the abscissa is always zero, so $x=0$.
Therefore, the equation of the $y$-axis is $x=0$.

Q2. Passing through the point $(-4,3)$ with slope $\frac{1}{2}$.

Sol. With $x_1=-4$, $y_1=3$ and $m=\frac{1}{2}$, applying the point-slope form $y-y_1=m(x-x_1)$:
or
$
y-3=\frac{1}{2}(x+4)
$
or
$
2 y-6=x+4
$
or
$
x-2 y+10=0 .
$

Q3. Passing through $(0,0)$ with slope $m$.

Sol. With $x_1=0$, $y_1=0$, the point-slope form gives $y-0=m(x-0)$,
or
$
y=m x .
$

Q4. Passing through $(2,2 \sqrt{3})$ and inclined with the $\boldsymbol{x}$-axis at an angle of $\mathbf{7 5}^{\circ}$.

Sol. Here $m$, the slope of line $=\tan 75^{\circ}=\tan \left(45^{\circ}+30^{\circ}\right)$

$
\begin{aligned}
& =\frac{\tan 45^{\circ}+\tan 30^{\circ}}{1-\tan 45^{\circ} \tan 30^{\circ}} \\
& =\frac{1+\frac{1}{\sqrt{3}}}{1-\frac{1}{\sqrt{3}}}=\frac{\frac{\sqrt{3}+1}{\sqrt{3}}}{\frac{\sqrt{3}-1}{\sqrt{3}}}=\frac{\sqrt{3}+1}{\sqrt{3}-1}
\end{aligned}
$

Also,
$
x_1=2, y_1=2 \sqrt{3}
$

Using point-slope form, equation of line is
$
y-2 \sqrt{3}=\frac{\sqrt{3}+1}{\sqrt{3}-1}(x-2)
$
Cross-multiplying,
or $\quad(\sqrt{3}-1) y-6+2 \sqrt{3}=(\sqrt{3}+1) x-2 \sqrt{3}-2$
or $\quad 4 \sqrt{3}-4=(\sqrt{3}+1) x-(\sqrt{3}-1) y$ or $(\sqrt{3}+1) x-(\sqrt{3}-1) y=4(\sqrt{3}-1)$.

Q5. Intersecting the $x$-axis at a distance of 3 units to the left of origin with slope – 2.

Sol. The line passes through $(-3,0)$ (3 units left of origin on the $x$-axis) with slope $-2$.
∴ Equation of line is $y-0=-2(x-(-3))$
or $\quad y=-2(x+3) \quad$ or $\quad y=-2 x-6$
or $\quad 2 x+y+6=0$.

Q6. Intersecting the $y$-axis at a distance of 2 units above the origin and making an angle of $30^{\circ}$ with positive direction of the $\boldsymbol{x}$-axis.

Sol. The line passes through $(0,2)$ (2 units above origin on $y$-axis) and has slope
$
=\tan 30^{\circ}=\frac{1}{\sqrt{3}}
$
∴ Equation of line is $y-2=\frac{1}{\sqrt{3}}(x-0)$
or $\quad \sqrt{3} y-2 \sqrt{3}=x \quad$ or $\quad x-\sqrt{3} y+2 \sqrt{3}=0$.

Q7. Passing through the points ( $-1,1$ ) and ( $2,-4$ ).

Sol. With $x_1=-1$, $y_1=1$, $x_2=2$, $y_2=-4$, using the two-point form:
$
y-y_1=\frac{y_2-y_1}{x_2-x_1}\left(x-x_1\right)
$

or
$
\begin{aligned}
y-1 & =\frac{-4-1}{2-(-1)}(x-(-1)) \text { or } y-1=\frac{-5}{3}(x+1) \\
3 y-3 & =-5 x-5 \quad \text { or } \quad 5 x+3 y+2 \\
3 & =0
\end{aligned}
$

Q8. Perpendicular distance from the origin is 5 units and the angle made by the perpendicular with the positive $\boldsymbol{x}$-axis is $\mathbf{3 0}^{\boldsymbol{\circ}}$.

Sol. Given $p=5$ and $\omega=30^{\circ}$, the normal form of the line is
$
\begin{aligned}
x \cos \omega+y \sin \omega & =p \\
x \cos 30^{\circ}+y \sin 30^{\circ} & =5 \quad \text { or } \quad x \cdot \frac{\sqrt{3}}{2}+y \cdot \frac{1}{2}=5
\end{aligned}
$

Multiplying by $2, \sqrt{3} x+y=10$.

Q9. The vertices of $\triangle \mathbf{P Q R}$ are $\mathbf{P}(\mathbf{2}, \mathbf{1}), \mathbf{Q}(-\mathbf{2}, 3)$ and $\mathbf{R}(4,5)$. Find equation of the median through the vertex $R$.

Sol. Let M be the midpoint of PQ:
$
\mathrm{M}=\left(\frac{2-2}{2}, \frac{1+3}{2}\right)=(0,2)
$
So RM is a median of $\triangle\mathrm{PQR}$.
Using the two-point form, the equation of median RM is
$
y-5=\frac{2-5}{0-4}(x-4)
$
(Here $x_1=4, y_1=5, x_2=0, y_2=2$ )
or $\quad y-5=\frac{3}{4}(x-4) \quad$ or $\quad 4 y-20=3 x-12$
or $3 x-4 y+8=0$.

Q10. Find the equation of the line passing through ( $-\mathbf{3}, \mathbf{5}$ ) and perpendicular to the line through the points $(2,5)$ and (-3, 6).

Sol. Let line $l_1$ pass through $\mathrm{A}(2,5)$ and $\mathrm{B}(-3,6)$.
Slope of $l_1=\frac{6-5}{-3-2}=-\frac{1}{5}$.
The required line $l_2$ is perpendicular to $l_1$.

∴ Slope of $l_2=5$
[- ve reciprocal]
The equation of $l_2$ through $\mathrm{P}(-3,5)$ with slope 5 is
or
$
\begin{array}{lrl}
y-5 & =5(x-(-3)) & \\
y-5 & \text { [Point-slope form] } \\
y-5 x+15 & \text { or } & 5 x-y+20=0
\end{array}
$

Q11. A line perpendicular to the line segment joining the points $(1,0)$ and $(2,3)$ divides it in the ratio $1: n$. Find the equation of the line.

Sol. The slope of segment $\mathrm{A}(1,0)$ to $\mathrm{B}(2,3)$ is $\frac{3-0}{2-1}=3$.

The required line $l$ is perpendicular to AB.
∴ Slope of $l=-\frac{1}{3}$
[- ve reciprocal]
Since $l$ divides AB internally in ratio $1:n$ at point P, by the section formula:
$\therefore \quad \mathrm{P}=\left(\frac{1 \times 2+n \times 1}{1+n}, \frac{1 \times 3+n \times 0}{1+n}\right)=\left(\frac{2+n}{1+n}, \frac{3}{1+n}\right)$
Using the point-slope form, the equation of $l$ is
$
y-\frac{3}{1+n}=-\frac{1}{3}\left(x-\frac{2+n}{1+n}\right)
$
Cross-multiplying,
or
$
3 y-\frac{9}{1+n}=-x+\frac{2+n}{1+n}
$
or
$
3(1+n) y-9=-(1+n) x+2+n
$
or
$
(1+n) x+3(1+n) y=n+11 .
$

Q12. Find the equation of a line that cuts off equal intercepts on the coordinate axes and passes through the point ( 2,3 ).

Sol. Let both intercepts equal $a$.
Then equation of line is
$
\frac{x}{a}+\frac{y}{a}=1
$
[Intercept form]
or
$
x+y=a
$

Since the line passes through $(2,3)$:
$
2+3=a \text { or } a=5
$

Putting $a=5$ in ( $i$ ), the equation of line is $x+y=5$.

Q13. Find equation of the line passing through the point ( 2,2 ) and cutting off intercepts on the axes whose sum is 9.

Sol. Let the $x$-intercept be $a$; then the $y$-intercept is
$
\begin{aligned}
b= & 9-a . \\
& {[\because \quad a+b=9 \text { (given) }] }
\end{aligned}
$

The equation of the line in intercept form is
$
\frac{x}{a}+\frac{y}{9-a}=1
$
…(i) $\quad\left|\frac{x}{a}+\frac{y}{b}=1\right|$
Since, it passes through the point ( 2,2 ), we have
$
\frac{2}{a}+\frac{2}{9-a}=1 \Rightarrow \frac{2(9-a)+2 a}{a(9-a)}=1
$
Cross-multiplying, we get
$
\begin{array}{rlrl}
& & 2(9-a)+2 a & =a(9-a) \\
\text { or } & 18-2 a+2 \mathrm{a} & =9 a-a^2 \\
\text { or } & (a-3)(a-6) & =0 \\
\therefore & & a & =3,6
\end{array} \quad \text { or } a^2-9 a+18=0
$

When $a=3$, from ( $i$ )
$
\frac{x}{3}+\frac{y}{}=1 \Rightarrow \frac{2 x+y}{6}=1
$
or
$
2 x+y=6 \text { or } 2 x+y-6=0
$

When $a=6$, from ( $i$ )
$
\frac{x}{3}+\frac{y}{3}=1 \Rightarrow \frac{x+2 y}{6}=1
$
or
$
x+2 y=6 \text { or } x+2 y-6=0
$

Hence, the equations of the line are
$
2 x+y-6=0, x+2 y-6=0
$

Q14. Find equation of the line through the point $(0,2)$ making an angle $\frac{2 \pi}{3}$ with the positive $x$-axis. Also, find the equation of line parallel to it and crossing the $y$-axis at a distance of 2 units below the origin.

Sol. The line passes through $(0,2)$.
$
\text { Its slope }=\tan \frac{2 \pi}{3}=\tan \left(\pi-\frac{\pi}{3}\right)=-\tan \frac{\pi}{3}=-\sqrt{3}
$
∴ Its equation is $y-2=-\sqrt{3}(x-0) \mid \because \boldsymbol{y}-\boldsymbol{y}_1=\boldsymbol{m}\left(\boldsymbol{x}-\boldsymbol{x}_1\right)$
or
$
y=-\sqrt{3} x+2
$

Any line parallel to (i) also has slope $-\sqrt{3}$. Cutting the $y$-axis 2 units below the origin means passing through $(0,-2)$, giving
$
y+2=-\sqrt{3}(x-0) \quad \text { or } y+2=-\sqrt{3} x \text { or } \quad \sqrt{3} x+y+2=0
$

Therefore, the required equations are $\sqrt{3} x+y-2=0$ and $\sqrt{3} x+y+2=0$.

Q15. The perpendicular from the origin to a line meets it at the point ( $-2,9$ ), find the equation of the line.

Sol. Slope of $\mathrm{OA} = \frac{9-0}{-2-0} = -\frac{9}{2}$. Since $\mathrm{OA}\perp l$, slope of $l = \frac{2}{9}$.

Using the point-slope form, the equation of $l$ is
$
y-9=\frac{2}{9}(x-(-2))
$
Cross-multiplying, $9(y-9)=2(x+2)$
or $\quad 9 y-81=2 x+4 \quad$ or $\quad 2 x-9 y+85=0$.

Q16. The length $L$ (in centimetre) of a copper rod is a linear function of its Celsius temperature $C$. In an experiment, if $L=124.942$ when $C=20$ and $L=125.134$ when $C=110$, express $L$ in terms of $C$.

Sol. Placing C on the $x$-axis and L on the $y$-axis, the two data points are $(20, 124.942)$ and $(110, 125.134)$. By the two-point form, any point $(\mathrm{C}, \mathrm{L})$ on the line satisfies
$
\begin{gathered}
\mathrm{L}-124.942=\frac{125.134-124.942}{110-20}(\mathrm{C}-20) \\
{\left[y-y_1=\frac{y_2-y_1}{x_2-x_1}\left(x-x_1\right)\right]}
\end{gathered}
$

or
$
\mathrm{L}-124.942=\frac{0.192}{90}(\mathrm{C}-20)
$
or
$
\mathrm{L}=\frac{0.192}{90}(\mathrm{C}-20)+124.942
$

This expresses $L$ as a linear function of $C$.

Q17. The owner of a milk store finds that, he can sell 980 litres of milk each week at $₹ 14 /$ litre and 1220 litres of milk each week at $₹ 16 /$ litre . Assuming a linear relationship between selling price and demand, how many litres could he sell weekly at $₹ 17 /$ litre?

Sol. Taking price (₹/litre) along the $x$-axis and demand (litres) along the $y$-axis, we use the two data points $(14,980)$ and $(16,1220)$.
Applying the two-point form to model the linear demand–price relationship:
$
\begin{aligned}
\mathrm{D}-980 & =\frac{1220-980}{16-14}(\mathrm{P}-14) \quad \text { । } y-y_1=\frac{y_2-y_1}{x_2-x_1}\left(x-x_1\right) \\
\text { or } \quad \mathrm{D} & =980+\frac{240}{2}(\mathrm{P}-14) \text { or } \mathrm{D}=980+120 \mathrm{P}-1680 \\
\text { or } \quad \mathrm{D} & =120 \mathrm{P}-700
\end{aligned}
$

Equation (i) gives demand D as a function of price P.
When $\mathrm{P}=17$, equation (i) gives $\mathrm{D}=120 \times 17-700$
or $\quad \mathrm{D}=1340$
Therefore, the owner can sell 1340 litres weekly at ₹17/litre.

Q18. $P(a, b)$ is the mid-point of $a$ line segment between axes. Show that equation of the line is $\frac{x}{a}+\frac{y}{b}=2$.

Sol. Write the line in intercept form:
$
\frac{x}{p}+\frac{y}{q}=1
$

This line meets the $x$-axis at $\mathrm{A}(p,0)$ and the $y$-axis at $\mathrm{B}(0,q)$. Since $\mathrm{P}(a,b)$ is the midpoint of AB:

$
\begin{aligned}
\left(\frac{p+0}{2}, \frac{0+q}{2}\right) & =(a, b) \\
\Rightarrow \quad \frac{p}{2}=a \quad \text { and } \quad \frac{q}{2} & =b \\
\Rightarrow \quad p=2 a \quad \text { and } \quad q & =2 b
\end{aligned}
$

Substituting back into (i), the equation of the line becomes
$
\frac{x}{2 a}+\frac{y}{2 b}=1 \text { multiplying every term by } 2, \frac{x}{a}+\frac{y}{b}=2 .
$

Q19. Point $\mathbf{R}(\boldsymbol{h}, \boldsymbol{k})$ divides a line segment between the axes in the ratio $1: 2$. Find equation of the line.

Sol. Write the line in intercept form:
$
\frac{x}{a}+\frac{y}{b}=1
$

If it meets the $x$-axis in A and $y$-axis in B , then coordinates of A and B are $(a, 0)$ and ( $0, b$ ) respectively.
Since $\mathrm{R}(h,k)$ divides AB in ratio $1:2$, the section formula gives
$
\begin{aligned}
& \left(\frac{1 \times 0+2 a}{1+2}, \frac{1 b+2 \times 0}{1+2}\right)=(h, k) \\
\Rightarrow & \frac{2 a}{3}=h \quad \text { and } \quad \frac{b}{3}=k \Rightarrow 2 a=3 h \text { and } b=3 k \\
\Rightarrow & a=\frac{3 h}{2} \quad \text { and } \quad b=3 k
\end{aligned}
$

Substituting $a$ and $b$ into (i), the equation becomes
$
\frac{x}{3 h / 2}+\frac{y}{3 k}=1 \quad \text { or } \quad \frac{2 x}{3 h}+\frac{y}{3 k}=1
$

Multiplying by $3 h k$, we have
$
2 k x+h y=3 h k .
$

Q20. By using the concept of equation of a line, prove that the three points $(3,0),(-2,-2)$ and $(8,2)$ are collinear.

Sol. The given points are $\mathrm{A}(3,0)$, $\mathrm{B}(-2,-2)$ and $\mathrm{C}(8,2)$.

The two-point form gives the line through $\mathrm{A}(3,0)$ and $\mathrm{B}(-2,-2)$ as
$
y-0=\frac{-2-0}{-2-3}(x-3)
$
or
$
y=\frac{2}{5}(x-3) \text { or } 5 y=2 x-6
$
or $2 x-5 y-6=0$

The three points are collinear if and only if $\mathrm{C}(8,2)$ satisfies equation (i).
Substituting $x=8$ and $y=2$ into (i):
$
2 \times 8-5 \times 2-6=0
$
or $\quad 16-10-6=0$ or $0=0$ which is true.
Since $0=0$ is true, the three points are collinear.

Exercise 9.3

Q1. Reduce the following equations into slope-intercept form and find their slopes and the $y$-intercepts. (i) $\boldsymbol{x}+\mathbf{7 y}=\mathbf{0}$, (ii) $\mathbf{6 x}+\mathbf{3 y}-\mathbf{5}=\mathbf{0}$, (iii) $\boldsymbol{y}=\mathbf{0}$.

Sol. (i) Rewrite $x+7y=0$ as
$
7 y=-x \quad \text { or } \quad y=-\frac{1}{7} x+0
$

Comparing with the standard slope-intercept form $y=mx+c$:
$
\text { slope } m=-\frac{1}{7}, y \text {-intercept } c=0
$
(ii) Given equation $6 x+3 y-5=0$ can be written as $3 y=-6 x+5$
Dividing by $3, y=-2 x+\frac{5}{3}$
Comparing with the standard slope-intercept form $y=mx+c$:
slope $m=-2, y$-intercept $c=\frac{5}{3}$.
(iii) The equation $y=0$ can be written as
$
y=0 x+0
$

Comparing with the standard slope-intercept form $y=mx+c$: slope $m=0, y$-intercept $c=0$.

Q2. Reduce the following equations into intercept form and find their intercepts on the axes.(i) $3 x+2 y-12=0$, (ii) $4 x-3 y=6$, (iii) $\mathbf{3 y} \boldsymbol{+} \mathbf{2} \boldsymbol{=} \mathbf{0}$.

Sol. (i) Rewrite $3x+2y-12=0$ as
$
3 x+2 y=12
$

Divide by 12 to get the intercept form (RHS $=1$):
$
\text { or } \frac{3 x}{12}+\frac{2 y}{12}=1 \text { or } \frac{x}{4}+\frac{y}{=1}=1
$

Matching with $\frac{x}{a}+\frac{y}{b}=1$:
$x$-intercept $\quad a=4, y$-intercept $b=6$.
(ii) Divide $4x-3y=6$ by 6:
$
\begin{aligned}
& \qquad \frac{4 x}{2}-\frac{3 y}{2}=1 \\
& \text { or } \frac{2 x}{3}-\frac{y}{2}=1 \\
& \text { or } \frac{x}{3 / 2}+\frac{y}{-2}=1 \\
& \text { Comparing with } \frac{x}{a}+\frac{y}{b}=1 \text {, we have } \\
& x \text {-intercept } a=\frac{3}{2}, y \text {-intercept } b=-2 \text {. }
\end{aligned}
$
(iii) Rewrite $3y+2=0$:
$
3 y=-2 \quad \text { or } \quad y=-\frac{2}{3}
$

This line is parallel to the $x$-axis, so it has no $x$-intercept.
Intercept with $y$-axis $=-\frac{2}{3}$.

Q3. Reduce the following equations into normal form. Find their perpendicular distances from the origin and angle between perpendicular and the positive $\boldsymbol{x}$-axis. (i) $x-\sqrt{3} y+8=0, \quad$ (ii) $y-2=0, \quad$ (iii) $x-y=4$.

Sol. (i) Rearrange to $-x+\sqrt{3}y=8$ (making the RHS positive).
Divide by $\sqrt{(-1)^2+(\sqrt{3})^2}=2$:

$
-\frac{1}{2} x+\frac{\sqrt{3}}{2} y=4
$

Comparing with $x \cos \omega+y \sin \omega=p \ldots(i)$, we get
$
\cos \omega=-\frac{1}{2}, \sin \omega=\frac{\sqrt{3}}{2}, p=4
$

Since $\cos\omega<0$ and $\sin\omega>0$, angle $\omega$ is in the second quadrant.
$
\begin{aligned}
& \cos \omega=-\cos 60^{\circ}=\cos \left(180^{\circ}-60^{\circ}\right)=\cos 120^{\circ} \\
& \Rightarrow \quad \omega=120^{\circ}
\end{aligned}
$

Putting values of $p$ and $\omega$ in ( $i$ ), $x \cos 120^{\circ}+y \sin 120^{\circ}=4$ is the normal form and $p=4, \omega=120^{\circ}$.
Remark: The interested reader, if he/she likes can find $\omega$ from $\sin \omega=\frac{\sqrt{3}}{2}$
(ii) Write $y-2=0$ as $0x+1y=2$ (RHS already positive).

Divide by $\sqrt{0^2+1^2}=1$ (no change):
$
0 x+1 y=2
$

Comparing with $x \cos \omega+y \sin \omega=p \ldots(i)$, we get $\cos \omega=0, \sin \omega=1, \quad p=2$
$
\Rightarrow \quad \omega=90^{\circ}
$

Putting values of $p$ and $\omega$ in ( $i$ ),
$x \cos 90^{\circ}+y \sin 90^{\circ}=2$ is the normal form and $p=2, \omega=90^{\circ}$.
(iii) The equation $x-y=4$ already has a positive RHS.

Divide by $\sqrt{1^2+(-1)^2}=\sqrt{2}$:
$
\frac{1}{\sqrt{2}} x-\frac{1}{\sqrt{2}} y=2 \sqrt{2}
$

Comparing with $x \cos \omega+y \sin \omega=p \ldots(i)$, we get
$
\cos \omega=\frac{1}{\sqrt{2}}, \sin \omega=-\frac{1}{\sqrt{2}}, p=2 \sqrt{2}
$
Since $\cos\omega>0$ and $\sin\omega<0$, $\omega$ is in the fourth quadrant.
$
\begin{aligned}
& \cos \omega=\cos 45^{\circ}=\cos \left(360^{\circ}-45^{\circ}\right)=\cos 315^{\circ} \\
& \Rightarrow \omega=315^{\circ}
\end{aligned}
$

Putting values of $p$ and $\omega$ in ( $i$ ),

$x \cos 315^{\circ}+y \sin 315^{\circ}=2 \sqrt{2}$ is the normal form and $p=2 \sqrt{2}, \omega=315^{\circ}$.

Q4. Find the distance of the point ( $-1,1$ ) from the line $12(x+6)=5(y-2)$.

Sol. Expand the given line $12(x+6)=5(y-2)$:
or
$
12 x+72=5 y-10
$
or
$
12 x-5 y+82=0
$
The perpendicular distance from $(-1,1)$ to this line is
$
\begin{aligned}
& \left.=\frac{|12(-1)-5(1)+82|}{\sqrt{(12)^2+(-5)^2}} \right\rvert\, \frac{\left|a x_1+b y_1+c\right|}{\sqrt{a^2+b^2}} \\
& =\frac{|-12-5+82|}{\sqrt{144+25}}=\frac{65}{13}=5
\end{aligned}
$

Q5. Find the points on the $\boldsymbol{x}$-axis, whose distances from the line $\frac{x}{3}+\frac{y}{4}=1$ are 4 units.

Sol. Let $\mathrm{P}(x, 0)$ be the required point on the $x$-axis.

Equation of the given line is $\frac{x}{3}+\frac{y}{4}=1$
or $4 x+3 y=12$ or $4 x+3 y-12=0$

The distance from $\mathrm{P}(x,0)$ to line (i) equals 4, so:
$
\left.\therefore \quad \frac{|4 x+3(0)-12|}{\sqrt{(4)^2+(3)^2}}=4 \quad \right\rvert\, \frac{\left|a x_1+b y_1+c\right|}{\sqrt{a^2+b^2}}
$
or $\quad \frac{|4 x-12|}{5}=4 \quad$ or $\quad|4 x-12|=20$
$\therefore \quad 4 x-12= \pm 20 \quad \mid \because$ If $|x|=a$ and $a \geq 0$; then $x= \pm a$
Dividing by 4
$
x-3= \pm 5 \quad \text { or } \quad x=3 \pm 5=8,-2
$
∴ Required points are $\mathrm{P}(x, 0)=\mathrm{P}(8,0)$ and $\mathrm{P}(-2,0)$.

Q6. Find the distance between parallel lines (i) $15 x+8 y-34=0$ and $15 x+8 y+31=0$ (ii) $\boldsymbol{l}(\boldsymbol{x}+\boldsymbol{y})+\boldsymbol{p}=\mathbf{0}$ and $\boldsymbol{l}(\boldsymbol{x}+\boldsymbol{y})-\boldsymbol{r}=\mathbf{0}$.

Sol. (i) Identifying $A=15$, $B=8$, $C_1=-34$, $C_2=31$, the distance between the parallel lines is
∴ Required distance is
$
d=\frac{\left|\mathrm{C}_1-\mathrm{C}_2\right|}{\sqrt{\mathrm{A}^2+\mathrm{B}^2}}=\frac{|-34-31|}{\sqrt{(15)^2+(8)^2}}=\frac{|-65|}{\sqrt{289}}=\frac{65}{17}
$

(ii) Here $A=l$, $B=l$, $C_1=p$, $C_2=-r$, so the distance is
∴ Required distance is
$
\begin{aligned}
d & =\frac{\left|\mathrm{C}_1-\mathrm{C}_2\right|}{\sqrt{\mathrm{A}^2+\mathrm{B}^2}}=\frac{|p-(-r)|}{\sqrt{l^2+l^2}}=\frac{|p+r|}{\sqrt{2 l^2}} & \\
& =\frac{|p+r|}{\sqrt{2}|l|} & {\left[\because \sqrt{l^2}=|l|\right] } \\
& =\frac{1}{\sqrt{2}}\left|\frac{p+r}{l}\right| . & {\left[\because \frac{|x|}{|y|}=\left|\frac{x}{y}\right|\right] }
\end{aligned}
$

Q7. Find equation of the line parallel to the line $3 x-4 y+2=0$ and passing through the point ( $-2,3$ ).

Sol. The given line is $3x-4y+2=0$ with slope $-\frac{a}{b}=-\frac{3}{-4}=\frac{3}{4}$.
$
\left[m=-\frac{a}{b}\right]
$

The required line is parallel to this, so it has the same slope $\frac{3}{4}$.
$
\left[m_1=m_2\right]
$

It passes through $(-2,3)$.
Therefore, the equation of the required line is
$
\begin{aligned}
y-3 & =\frac{3}{4}[x-(-2)] \quad \text { [Point-slope form] } \\
\text { or } \quad 4 y-12 & =3 x+6 \\
\text { or } \quad 3 x-4 y+18 & =0
\end{aligned}
$

Q8. Find equation of the line perpendicular to the line $x-7 y+5=0$ and having $x$-intercept 3 .

Sol. The given line $x-7y+5=0$ has slope $-\frac{a}{b}=-\frac{1}{-7}=\frac{1}{7}$.
$
\left[m=-\frac{a}{b}\right]
$

The required line is perpendicular to this, so its slope is $-7$.
[- ve reciprocal]
Since the $x$-intercept is 3, the line passes through $(3,0)$.
The equation of the line with slope $-7$ through $(3,0)$ is
or
$
\begin{aligned}
y-0 & =-7(x-3) & & \text { [Point-slope form] } \\
y & =-7 x+21 & \text { or } & 7 x+y=21 .
\end{aligned}
$

Q9. Find angles between the lines $\sqrt{3} x+y=1$ and $\boldsymbol{x}+\sqrt{\mathbf{3}} \boldsymbol{y}=\mathbf{1}$.

Sol. The slopes of the two lines are
$
m_1=\frac{-a}{b}=-\frac{\sqrt{3}}{1}=-\sqrt{3} \quad \text { and } \quad m_2=-\frac{1}{\sqrt{3}}
$

The acute angle $\theta$ between them satisfies
$
\begin{aligned}
\tan \theta & =\left|\frac{m_1-m_2}{1+m_1 m_2}\right|=\left|\frac{-\sqrt{3}+\frac{1}{\sqrt{3}}}{1+(-\sqrt{3})\left(-\frac{1}{\sqrt{3}}\right)}\right| \\
\Rightarrow \tan \theta & =\left|\frac{\frac{-3+1}{\sqrt{3}}}{1+1}\right|=\frac{\frac{2}{\sqrt{3}}}{2}=\frac{1}{\sqrt{3}}=\tan 30^{\circ}
\end{aligned}
$
So $\theta=30^{\circ}$. The two lines meet at angles of $30^{\circ}$ and $150^{\circ}$.
Remark: We know that there are two angles between two lines $\theta$ and $180^{\circ}-\theta$ (linear pair).

Q10. The line through the points $(h, 3)$ and $(4,1)$ intersects the line $7 x-9 y-19=0$ at right angles. Find the value of $\boldsymbol{h}$.

Sol. The slope of line $\mathrm{L}_1$ through $(h,3)$ and $(4,1)$ is
$
m_1=\frac{1-3}{4-h}=\frac{-2}{4-h}
$

The slope of line $\mathrm{L}_2: 7x-9y-19=0$ is
$
m_2=-\frac{a}{b}=-\frac{7}{-9}=\frac{7}{9}
$

For perpendicularity, $m_1 m_2=-1$:
$\Rightarrow\left(\frac{-2}{4-h}\right)\left(\frac{7}{9}\right)=-1$ or $\frac{-14}{36-9 h}=-1 \quad$ or $-14=-36+9 h$
or $\quad 9 h=22$
$\therefore \quad h=\frac{22}{9}$.

Q11. Prove that the line through the point ( $x_1, y_1$ ) and parallel to the line $\mathrm{Ax}+\mathrm{By}+\mathrm{C}=0$ is $ \mathbf{A}\left(x-x_1\right)+\mathbf{B}\left(y-y_1\right)=0 $

Sol. The given line $\mathrm{A}x+\mathrm{B}y+\mathrm{C}=0$ has slope $-\frac{A}{B}$. A parallel line has the same slope, so the equation through $\left(x_1,y_1\right)$ is
or
$
y-y_1=-\frac{\mathrm{A}}{\mathrm{~B}}\left(x-x_1\right)
$
or $\mathrm{A}\left(x-x_1\right)+\mathrm{B}\left(y-y_1\right)=0$.

Q12. Two lines passing through the point ( 2,3 ) intersect each other at an angle of $60^{\circ}$. If slope of one line is 2 ; find equation of the other line.

Sol. Slope of one line is $m_1=2$. Let the slope of the other line be $m_2$.
Since angle between the lines is $60^{\circ}$, we have
$
\begin{aligned}
\tan 60^{\circ} & =\left|\frac{m_1-m_2}{1+m_1 m_2}\right| \\
\text { or } \quad \sqrt{3} & =\left|\frac{2-m_2}{1+2 m_2}\right| \quad \text { or } \quad \frac{2-m_2}{1+2 m_2}= \pm \sqrt{3}
\end{aligned}
$

Taking the positive sign:, we have
$
\begin{aligned}
& 2-m_2 & =\sqrt{3}\left(1+2 m_2\right) \\
\text { or } & 2-m_2 & =\sqrt{3}+2 \sqrt{3} m_2 \\
\text { or } & (1+2 \sqrt{3}) m_2 & =2-\sqrt{3} \\
\therefore & m_2 & =\frac{2-\sqrt{3}}{1+2 \sqrt{3}}
\end{aligned}
$
∴ Equation of line through the point ( 2,3 ) and having slope $\frac{2-\sqrt{3}}{1+2 \sqrt{3}}$ is
$
\begin{aligned}
y-3 & =\frac{2-\sqrt{3}}{1+2 \sqrt{3}}(x-2) \\
\text { or } \quad(1+2 \sqrt{3}) y-3-6 \sqrt{3} & =(2-\sqrt{3}) x-4+2 \sqrt{3} \\
\text { or } \quad(\sqrt{3}-2) x+(1+2 \sqrt{3}) y & =-1+8 \sqrt{3}
\end{aligned}
$

Taking the negative sign:, we have
or
$
\begin{aligned}
& 2-m_2=-\sqrt{3}\left(1+2 m_2\right) \\
& 2-m_2=-\sqrt{3}-2 \sqrt{3} m_2 \\
& \sqrt{3}+2=(1-2 \sqrt{3}) m_2
\end{aligned}
$
or
$
\begin{array}{r}
\sqrt{3}+2=(1-2 \sqrt{3}) n \\
m_2=\frac{\sqrt{3}+2}{1-2 \sqrt{3}}
\end{array}
$
∴ Equation of line through the point $(2,3)$ and having slope $\frac{\sqrt{3}+2}{1-2 \sqrt{3}}$ is
$
y-3=\frac{\sqrt{3}+2}{1-2 \sqrt{3}}(x-2)
$
or $\quad(1-2 \sqrt{3}) y-3+6 \sqrt{3}=(\sqrt{3}+2) x-2 \sqrt{3}-4$
or $\quad(\sqrt{3}+2) x+(2 \sqrt{3}-1) y=8 \sqrt{3}+1$.

Q13. Find the equation of the right bisector of the line segment joining the points ( 3,4 ) and ( $-1,2$ ).

Sol. Let $\mathrm{A}(3,4)$ and $\mathrm{B}(-1,2)$ be the two given points.
Mid-point of AB is $\mathrm{M}=\left(\frac{3-1}{2}, \frac{4+2}{2}\right)=(1,3)$.
Right bisector of AB is the line $l$ through M and perpendicular to AB .
Slope of $\mathrm{AB}=\frac{2-4}{-1-3}=\frac{-2}{-4}=\frac{1}{2}$
Since $l \perp \mathrm{AB}$, slope of $l=-2$.
∴ Equation of right bisector $l$ is
$
y-3=-2(x-1)
$
or $\quad y-3=-2 x+2 \quad$ or $\quad 2 x+y=5$.

Q14. Find the coordinates of the foot of perpendicular from the point $(-1,3)$ to the line $3 x-4 y-16=0$.

Sol. Given line is AB : $3 x-4 y-16=0$.
Given point is $\mathrm{P}(-1,3)$.
Draw $\mathrm{PM} \perp \mathrm{AB}$. We have to find the coordinates of M .
Slope of $\mathrm{AB}=-\frac{3}{-4}=\frac{3}{4}$.

Since $\mathrm{PM} \perp \mathrm{AB}$, slope of $\mathrm{PM}=-\frac{4}{3} \quad(-$ ve reciprocal $)$
Using point-slope form, equation of PM is

$
y-3=-\frac{4}{3}(x-(-1))
$
or $\quad 3 y-9=-4 x-4 \quad$ or $\quad 4 x+3 y-5=0$
Now AB: $3 x-4 y-16=0$

PM: $4 x+3 y-5=0$

Now foot M of perpendicular is the point of intersection of lines (i) and (ii). So let us solve them for $\boldsymbol{x}$ and $\boldsymbol{y}$.
$\operatorname{Eqn}(i) \times 3+\operatorname{Eqn}(i i) \times 4$ gives to eliminate y ,
$9 x-12 y-48+16 x+12 y-20=0$
$\Rightarrow 25 x-68=0 \Rightarrow 25 x=68 \Rightarrow x=\frac{68}{25}$.
Putting $x=\frac{68}{25}$ in (i), $\frac{204}{25}-4 y-16=0$
$\Rightarrow-4 y=16-\frac{204}{25}=\frac{400-204}{25}=\frac{196}{25}$
$\therefore y=\frac{-196}{4 \times 25}=\frac{-49}{25}$
∴ co-ordinates of foot of perpendicular M are $\mathrm{M}\left(\frac{68}{25}, \frac{-49}{25}\right)$.

Q15. The perpendicular from the origin to the line $y=m x+c$ meets it at the point (-1,2). Find the values of $m$ and $c$.

Sol. Let OA be perpendicular from the origin $\mathrm{O}(0,0)$ to the line $l$ whose equation is $y=m x+c$. (slope – intercept form)

Slope of line $l=m$.
Slope of $\mathrm{OA}=\frac{2-0}{-1-0}=-2$.

Since $l \perp \mathrm{OA}$, we have $(m)(-2)=-1 \quad\left|m_1 m_2=-1\right|$
$\Rightarrow-2 m=-1$ or $m=\frac{1}{2}$
∴ Equation of line $l$ becomes
$
y=\frac{1}{2} x+c
$

The point $\mathrm{A}(-1,2)$ lies on it
$
\therefore \quad 2=\frac{1}{2}(-1)+c \text { or } 2=\frac{-1}{2}+c \text { or } \quad c=2+\frac{1}{2}=\frac{5}{2}
$

Hence $\quad m=\frac{1}{2}, c=\frac{5}{2}$.

Q16. If $p$ and $q$ are the lengths of perpendiculars from the origin to the lines $x \cos \theta-y \sin \theta=k \cos 2 \theta$ and $x \sec \theta+y \operatorname{cosec} \theta=k$, respectively, prove that $p^2+4 q^2=k^2$.

Sol. Equations of the given lines are
$x \cos \theta-y \sin \theta=k \cos 2 \theta \quad$ or $\quad x \cos \theta-y \sin \theta-k \cos 2 \theta=0$
and $\quad x \sec \theta+y \operatorname{cosec} \theta=k$ or $x \sec \theta+y \operatorname{cosec} \theta-k=0 \ldots(i i) p=$ length of ⟂ from origin ( 0,0 ) on line ( $i$ )
$=\frac{\left|a x_1+b y_1+c\right|}{\sqrt{a^2+b^2}}$
$=\frac{|0-0-k \cos 2 \theta|}{\sqrt{\cos ^2 \theta+\sin ^2 \theta}}=|k \cos 2 \theta|$
$q=$ Length of ⟂ from origin ( 0,0 ) on line ( $i i$ )
$
\begin{aligned}
& =\frac{|0+0-k|}{\sqrt{\sec ^2 \theta+\operatorname{cosec}^2 \theta}}=\frac{|k|}{\sqrt{\frac{1}{\cos ^2 \theta}+\frac{1}{\sin ^2 \theta}}} \\
& =\frac{|k|}{\sqrt{\frac{\sin ^2 \theta+\cos ^2 \theta}{\cos ^2 \theta \sin ^2 \theta}}}
\end{aligned}
$
or $q=|k| \cos \theta \sin \theta$

Substituting the expressions for $p$ and $q$:
$
\begin{aligned}
\text { L.H.S. } & =k^2 \cos ^2 2 \theta+4 k^2 \cos ^2 \theta \sin ^2 \theta \\
& =k^2 \cos ^2 2 \theta+k^2(2 \sin \theta \cos \theta)^2 \\
& =k^2 \cos ^2 2 \theta+k^2 \sin ^2 2 \theta \\
& =k^2\left(\cos ^2 2 \theta+\sin ^2 2 \theta\right)=k^2=\text { R.H.S. }
\end{aligned}
$

Q17. In the triangle $\mathbf{A B C}$ with vertices $\mathbf{A}(\mathbf{2}, \mathbf{3}), \mathbf{B}(\mathbf{4},-\mathbf{1})$ and $C(1,2)$, find the equation and length of altitude from the vertex $A$.

Sol. Slope of $\mathrm{BC}=\frac{2-(-1)}{1-4}=-1$. Drawing $\mathrm{AD}\perp\mathrm{BC}$, the altitude AD has slope $1$ (negative reciprocal). (-ve reciprocal)

Using point-slope form, equation of altitude AD is
$
y-3=1(x-2)
$

Equation of BC is
$
y-(-1)=\frac{2-(-1)}{1-4}(x-4)
$
[Two-point form]
$
\begin{aligned}
& \text { or } y+1=\frac{3}{-3}(x-4) \quad \text { or } \quad y+1=-(x-4) \\
& \text { or } y+1=-x+4 \\
& \text { or } \quad x+y-3=0
\end{aligned}
$

Length of altitude AD
$
\begin{aligned}
& =\text { Length of perpendicular form } \mathrm{A}(2,3) \text { on } \mathrm{BC} \\
& =\frac{|2+3-3|}{\sqrt{1^2+1^2}} \\
& =\frac{2}{\sqrt{2}}=\sqrt{2} .
\end{aligned}
$

Q18. If $p$ is the length of perpendicular from the origin to the line whose intercepts on the axes are $a$ and $b$, then show that $\frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}$.

Sol. A line with axis-intercepts $a$ and $b$ has equation $\frac{x}{a}+\frac{y}{b}-1=0$. The perpendicular distance from the origin to this line is $p$ (given), so
$
\begin{aligned}
& \therefore \quad p=\frac{|0+0-1|}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}}=\frac{1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}} \\
& \qquad p^2=\frac{1}{\frac{1}{a^2}+\frac{1}{b^2}} \\
& \text { Squaring both sides, } \\
& \text { Cross-multiplying } p^2\left(\frac{1}{a^2}+\frac{1}{b^2}\right)=1 \\
& \text { Dividing by } p^2, \quad \frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{p^2} . \\
& \text { Or } \frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2} .
\end{aligned}
$

Miscellaneous Exercise

Q1. Find the values of $\boldsymbol{k}$ for which the line $(k-3) x-\left(4-k^2\right) y+k^2-7 k+6=0$ is (i) parallel to the $\boldsymbol{x}$-axis (ii) parallel to the $\boldsymbol{y}$-axis (iii) passing through the origin.

Sol. The given line is $(k-3)x-(4-k^2)y+k^2-7k+6=0$.
(i) For the line to be parallel to the $x$-axis, it must have the form $y=b$, meaning the coefficient of $x$ must vanish:
$
\therefore \quad k-3=0 \text { or } k=3 .
$
(ii) For the line to be parallel to the $y$-axis, it must have the form $x=a$, so the coefficient of $y$ must be zero:
$
\therefore \quad-\left(4-k^2\right)=0 \text { or }-4+k^2=0 \text { or } k^2=4 \quad \therefore k= \pm 2 \text {. }
$
(iii) For the line to pass through the origin $(0,0)$, substituting $x=0$, $y=0$ into (i): \\
\text { i.e., } 0-0+k^2-7 k+6=0 \\
\text { i.e., } \\
(k-1)(k-6)=0 \quad \therefore \quad k=6 \text { or } 1 .
\end{gathered}
$

Q2. Find the values of $\theta$ and $p$, if the equation $x \cos \theta+y \sin \theta=p$ is the normal form of the line $\sqrt{3} \boldsymbol{x}+\boldsymbol{y}+\mathbf{2}=\mathbf{0}$.

Sol. Rearrange: $\sqrt{3}x+y+2=0 \Rightarrow -\sqrt{3}x-y=2$ (RHS made positive by multiplying by $-1$).
Divide by $\sqrt{(-\sqrt{3})^2+(-1)^2}=2$:
$
-\frac{\sqrt{3}}{2} x-\frac{1}{2} y=1
$

Comparing with $x \cos \theta+y \sin \theta=p$, we get

$
p=1, \cos \theta=-\frac{\sqrt{3}}{2}, \sin \theta=-\frac{1}{2}
$

Since both $\cos\theta$ and $\sin\theta$ are negative, $\theta$ is in the third quadrant.
$
\begin{aligned}
& \begin{aligned}
\therefore \quad \cos \theta & =-\frac{\sqrt{3}}{2}=-\cos 30^{\circ} \\
& =\cos \left(180^{\circ}+30^{\circ}\right)=\cos 210^{\circ}
\end{aligned} \\
& \Rightarrow \quad \theta \\
& \text { Hence, } \theta \\
& =210^{\circ}=\frac{7 \pi}{} \text { and } p=1 .
\end{aligned}
$

Note: The interested reader may find $\theta$ from $\sin \theta=\frac{-1}{2}$.

Q3. Find the equations of the lines, which cut-off intercepts on the axes whose sum and product are 1 and – 6, respectively.

Sol. Let the $x$- and $y$-intercepts be $a$ and $b$, giving the intercept-form equation $\frac{x}{a}+\frac{y}{b}=1$.

The given conditions are: $a+b=1$ and $ab=-6$.

From the sum condition, $b=1-a$. Substituting into the product condition:

Putting this value of $b$ in equation (iii),
$
\begin{gathered}
a(1-a)=-6 \text { or } a-a^2+6=0 \\
\text { or }-a^2+a+6=0 \text { or } a^2-a-6=0 \text { or }(a-3)(a+2)=0
\end{gathered}
$
∴ Either $\quad a=3 \quad$ or $\quad a=-2$
If $a=3$, from equation (iv), $b=1-a=1-3=-2$
Putting these values of $a$ and $b$ in ( $i$ ), equation of one required line is
$
\frac{x}{3}+\frac{y}{-2}=1 \quad \text { or } \quad \frac{x}{3}-\frac{y}{2}=1 \text { or } \frac{2 x-3 y}{6}=1
$
or $\quad 2 x-3 y=6$
If $a=-2$, from equation (iv), $b=1-a=1+2=3$
Putting these values of $a$ and $b$ in ( $i$ ), equation of second required line is $\frac{x}{-2}+\frac{y}{3}=1$.
Multiplying by $-6,3 x-2 y=-6$ or $3 x-2 y+6=0$.

Q4. What are the points on the $\boldsymbol{y}$-axis whose distance from the line $\frac{x}{3}+\frac{y}{4}=1$ is 4 units?

Sol. Let $(0, y)$ be a point on the $y$-axis. Its distance from the line
$
\begin{aligned}
& \\
& \text { or } \\
& \quad \frac{x}{3}+\frac{y}{4}=1 \text { i.e. } \frac{4 x+3 y}{12}=1 \Rightarrow 4 x+3 y=12 \\
& \quad \frac{\mid 4 x+3 y-12}{\sqrt{4^2+3^2}}=0 \text { is } 4 \text { units, then } \\
& \Rightarrow \quad \frac{|3 y-12|}{5}=4 \\
& \Rightarrow \quad 3 y-12= \pm 20 \quad \Rightarrow \quad|3 y-12|=20 \\
& \Rightarrow \quad 3 y= \pm 20+12 \quad \Rightarrow 3 y=32 \text { or }-8 \\
& \Rightarrow \quad y=\frac{32}{3} \text { or }-\frac{8}{3}
\end{aligned}
$

Hence, the required points on $y$-axis are $(0, y)=\left(0, \frac{32}{3}\right)$ and $\left(0,-\frac{8}{3}\right)$.

Q5. Find perpendicular distance from the origin of the line joining the points $(\cos \theta, \sin \theta)$ and $(\cos \phi, \sin \phi)$.

Sol. Applying the two-point form, the line through $(\cos\theta, \sin\theta)$ and $(\cos\phi, \sin\phi)$ is
$
\begin{aligned}
& y-\sin \theta=\frac{\sin \phi-\sin \theta}{\cos \phi-\cos \theta}(x-\cos \theta) \\
& \text { or } \quad y-\sin \theta=\frac{2 \cos \frac{\phi+\theta}{2} \sin \frac{\phi-\theta}{2}}{-2 \sin \frac{\phi+\theta}{2} \sin \frac{\phi-\theta}{2}}(x-\cos \theta) \\
& \text { or } \quad y-\sin \theta=-\frac{\cos \frac{\phi+\theta}{2}}{\sin \frac{\phi+\theta}{2}}(x-\cos \theta) \\
& \text { Cross-multiplying } \\
& (y-\sin \theta) \sin \frac{\phi+\theta}{2}=-\cos \frac{\phi+\theta}{2}(x-\cos \theta) \\
& \text { or } \quad y \sin \frac{\phi+\theta}{2}-\sin \theta \sin \frac{\phi+\theta}{2} \\
& \quad=-x \cos \frac{\phi+\theta}{2}+\cos \theta \cos \frac{\phi+\theta}{2}
\end{aligned}
$

$
\begin{gathered}
\text { or } x \cos \frac{\phi+\theta}{2}+y \sin \frac{\phi+\theta}{2}- \\
\left(\cos \theta \cos \frac{\phi+\theta}{2}+\sin \theta \sin \frac{\phi+\theta}{2}\right)=0 \\
\text { or } x \cos \frac{\phi+\theta}{2}+y \sin \frac{\phi+\theta}{2}-\cos \left(\theta-\frac{\phi+\theta}{2}\right)=0 \\
{[\because \cos \mathrm{~A} \cos \mathrm{~B}+\sin \mathrm{A} \sin \mathrm{~B}=\cos (\mathrm{A}-\mathrm{B})]} \\
\text { or } \quad x \cos \frac{\phi+\theta}{2}+y \sin \frac{\phi+\theta}{2}-\cos \frac{\theta-\phi}{2}=0 \\
{\left[\because \theta-\left(\frac{\theta+\phi}{2}\right)=\frac{2 \theta-\theta-\phi}{2}=\frac{\theta-\phi}{2}\right]}
\end{gathered}
$

The perpendicular distance from the origin $(0,0)$ to this line is
$
=\frac{\left|0+0-\cos \frac{\theta-\phi}{2}\right|}{\sqrt{\cos ^2 \frac{\phi+\theta}{2}+\sin ^2 \frac{\phi+\theta}{2}}}=\left|\cos \frac{\theta-\phi}{2}\right|(\because|-t|=|t|)
$

Note 1. The answer may be written as
$
\begin{aligned}
& \left|\cos \frac{\phi-\theta}{2}\right| \\
= & \left|\frac{2 \sin \frac{\phi-\theta}{2} \cos \frac{\phi-\theta}{2}}{2 \sin \frac{\phi-\theta}{2}}\right| \\
= & \frac{|\sin (\phi-\theta)|}{\left|2 \sin \frac{\phi-\theta}{2}\right|} \quad[\because 2 \cos (-x)=\cos x] \\
= & \frac{|\sin (\phi-\theta)|}{2\left|\sin \frac{\phi-\theta}{2}\right|}
\end{aligned}
$

Alternative Approach:
Equation of line is
$
\begin{aligned}
y-\sin \theta & =\frac{\sin \phi-\sin \theta}{\cos \phi-\cos \theta}(x-\cos \theta) \\
\text { or } \quad y(\cos \phi & -\cos \theta)-\sin \theta \cos \phi+\sin \theta \cos \theta \\
& =x(\sin \phi-\sin \theta)-\sin \phi \cos \theta+\sin \theta \cos \theta
\end{aligned}
$

$
\text { or } \begin{aligned}
(\sin \phi-\sin \theta) x-(\cos \phi-\cos \theta) y-\sin \phi & \cos \theta \\
& +\cos \phi \sin \theta=0
\end{aligned}
$
or $\sin \phi-\sin \theta) x-(\cos \phi-\cos \theta) y-\sin (\phi-\theta)=0$
Its distance from origin ( 0,0 ) is
$
\begin{aligned}
& =\frac{|0-0-\sin (\phi-\theta)|}{\sqrt{(\sin \phi-\sin \theta)^2+(\cos \phi-\cos \theta)^2}} \\
& =\frac{|\sin (\phi-\theta)|}{\sqrt{\left(\sin ^2 \phi+\cos ^2 \phi\right)+\left(\sin ^2 \theta+\cos ^2 \theta\right)-2(\cos \phi \cos \theta+\sin \phi \sin \theta)}} \\
& =\frac{|\sin (\phi-\theta)|}{\sqrt{1+1-2 \cos (\phi-\theta)}}=\frac{|\sin (\phi-\theta)|}{\sqrt{2[1-\cos (\phi-\theta)]}} \\
& =\frac{|\sin (\phi-\theta)|}{\sqrt{2 \times 2 \sin ^2 \frac{\phi-\theta}{2}}}=\frac{|\sin (\phi-\theta)|}{2 \sqrt{\sin ^2 \frac{\phi-\theta}{2}}} \\
& =\frac{|\sin (\phi-\theta)|}{2\left|\sin \frac{\phi-\theta}{2}\right|}
\end{aligned}
$

Q6. Find the equation of the line parallel to $y$-axis and drawn through the point of intersection of the lines $x-7 y+5=0$ and $3 x+y=0$.

Sol. To find the intersection point of
$
x-7 y+5=0
$
and $\quad 3 x+y=0$
we solve (i) and (ii) simultaneously.
$
\begin{aligned}
& \text { eqn }(i)+7 \times \text { eqn(ii) gives (To eliminate } y \text { ), } \\
& x-7 y+5+21 x+7 y=0 \Rightarrow 22 x+5=0 \\
& \Rightarrow 22 x=-5 \Rightarrow x=-\frac{5}{22}
\end{aligned}
$

Putting $x=-\frac{5}{22}$ in (ii), $-\frac{15}{22}+y=0$ or $y=\frac{15}{22}$
∴ The point of intersection of given lines is
$
\left(-\frac{5}{22}, \frac{15}{22}\right)=\left(x_1, y_1\right) .
$

The line through this point and parallel to the $y$-axis has equation
$
x=x_1, \text { i.e. }, \quad x=-\frac{5}{22} .
$

Q7. Find the equation of a line drawn perpendicular to the line $\frac{x}{4}+\frac{y}{6}=1$ through the point where it meets the $\boldsymbol{y}$-axis.

Sol. Rewrite the given line as $3x+2y=12$.

Its slope $\quad=-\frac{3}{2}$

Required line $l$ is perpendicular to the given line.
∴ Slope of $l=\frac{2}{3}$.

(- ve reciprocal)

Line (i) meets the $y$-axis at $\mathrm{B}(0,6)$ (found by setting $x=0$).
(Because putting $x=0$ in ( $i$ ), $2 y=12 \Rightarrow y=6$ )
The required line $l$ passes through $\mathrm{B}(0,6)$ with slope $\frac{2}{3}$.
∴ Equation of $l$ is
[Point-slope form]
or
$3 y-18=2 x$
or $-2 x+3 y-18=0$
or $2 x-3 y+18=0$.

Q8. Find the area of the triangle formed by the lines $\boldsymbol{y}-\boldsymbol{x}=0, \boldsymbol{x}+\boldsymbol{y}=0$ and $\boldsymbol{x}-\boldsymbol{k}=0$.

Sol. Name the triangle ABC with its sides given by
BC:
$
y-x=0
$

CA:
$
x+y=0
$

AB :
$
x-k=0
$

Finding vertex A by solving (ii) and (iii):
From (iii), $x=k$.
Putting $x=k$ in (ii), $k+y=0 \therefore y=-k$
$\therefore \quad$ vertex A is $(k,-k)=\left(x_1, y_1\right)$ (say)
Finding vertex B by solving (i) and (iii):
From (iii), $x=k$
Putting $x=k$ in ( $i$ ), $y-k=0 . \therefore y=k$
$\therefore \quad$ vertex B is $(k, k)=\left(x_2, y_2\right)$ (say)
To find vertex $\mathbf{C}$, we solve (i) and (ii) simultaneously.

Adding (i) and (ii), $2 y=0$ or $y=\frac{0}{2}=0$.
Putting $y=0$ in (ii), $x=0$
$\therefore \quad$ vertex C is $(0,0)=\left(x_3, y_3\right)$ (say)
Here $x_1=k, y_1=-k ; x_2=k, y_2=k ; x_3=0, y_3=0$.
∴ Area of triangle ABC
$=\frac{1}{2}\left|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right|$
$=\frac{1}{2}|k(k-0)+k(0+k)+0(-k-k)|$
$=\frac{1}{2}\left|k^2+k^2\right|=\frac{1}{2}\left|2 k^2\right|=\frac{1}{2}\left(2 k^2\right)=k^2$ sq. units.

Q9. Find the value of $p$ so that the three lines $3 x+y-2=0, p x+2 y-3=0$ and $2 x-y-3=0$ may intersect at one point.

Sol. The three lines are
$
\begin{array}{r}
3 x+y-2=0 \\
p x+2 y-3=0 \\
2 x-y-3=0
\end{array}
$

Find the intersection of (i) and (iii) first (since $p$ does not appear in them):
Adding $5 x-5=0$ or $5 x=5$ or $x=1$.
Putting $\quad x=1$ in $(i), \quad 3+y-2=0$ or $y+1=0$ or $y=-1$
∴ Point of intersection of (i) and (iii) is ( $1,-1$ ).
For all three lines to be concurrent, this point must also lie on (ii):
$\therefore p(1)+2(-1)-3=0 \Rightarrow p-2-3=0 \Rightarrow p=5$.

Q10. If three lines whose equations are $y=m_1 x+c_1$, $y=m_2 x+c_2$ and $y=m_3 x+c_3$ are concurrent, then show that $ m_1\left(c_2-c_3\right)+m_2\left(c_3-c_1\right)+m_3\left(c_1-c_2\right)=0 . $

Sol. The equations of the three lines are
$
\begin{aligned}
& y=m_1 x+c_1 \\
& y=m_2 x+c_2 \\
& y=m_3 x+c_3
\end{aligned}
$

To find the point of intersection of any two lines, say (ii) and (iii).
Subtracting (iii) from (ii), $0=\left(m_2-m_3\right) x+c_2-c_3$
or
$
c_3-c_2=\left(m_2-m_3\right) x
$
$\therefore \quad x=\frac{c_3-c_2}{m_2-m_3}$

Putting in (ii),
$
\begin{aligned}
y & =m_2\left(\frac{c_3-c_2}{m_2-m_3}\right)+c_2=\frac{m_2\left(c_3-c_2\right)+\left(m_2-m_3\right) c_2}{m_2-m_3} \\
& =\frac{m_2 c_3-m_2 c_2+m_2 c_2-m_3 c_2}{m_2-m_3}=\frac{m_2 c_3-m_3 c_2}{m_2-m_3}
\end{aligned}
$
∴ Point of intersection of line (ii) and (iii) is
$
\mathrm{P}\left(\frac{c_3-c_2}{m_2-m_3}, \frac{m_2 c_3-m_3 c_2}{m_2-m_3}\right)
$

Since the three lines are concurrent, P lies on (i).
$
\begin{aligned}
\Rightarrow \quad & \frac{m_2 c_3-m_3 c_2}{m_2-m_3}=m_1\left(\frac{c_3-c_2}{m_2-m_3}\right)+c_1 \\
& =\frac{m_1\left(c_3-c_2\right)+c_1\left(m_2-m_3\right)}{m_2-m_3} \\
\Rightarrow \quad & m_2 c_3-m_3 c_2=m_1 c_3-m_1 c_2+m_2 c_1-m_3 c_1 \\
\Rightarrow \quad & m_1 c_2-m_1 c_3+m_2 c_3-m_2 c_1+m_3 c_1-m_3 c_2=0 \\
\Rightarrow \quad & m_1\left(c_2-c_3\right)+m_2\left(c_3-c_1\right)+m_3\left(c_1-c_2\right)=0 .
\end{aligned}
$

Q11. Find the equations of the lines through the point $(3,2)$ which make an angle of $45^{\circ}$ with the line $x-2 y=3$.

Sol. Given line is $l$ : $x-2 y=3$
Its slope $=\frac{1}{2}$
Equation of any line through $\mathrm{P}(3,2)$ is
$
y-2=m(x-3)
$
where $m$ is the slope of line.

Since the angle between (i) and (ii) is $45^{\circ}$.
We have
$
\begin{aligned}
& \tan 45^{\circ}=\left|\frac{m-\frac{1}{2}}{1+m \cdot \frac{1}{2}}\right|=\left|\frac{\frac{2 m-1}{2}}{\frac{2+m}{2}}\right|\left[\because \tan \theta=\left|\frac{m_1-m_2}{1+m_1 m_2}\right|\right] \\
& \Rightarrow \quad 1=\left|\frac{2 m-1}{m+2}\right| \Rightarrow \frac{2 m-1}{m+2}= \pm 1 . \\
& \text { Taking the positive sign:, } 2 m-1=m+2 \\
& \Rightarrow \quad m=3
\end{aligned}
$

Putting this value of $m$ in (ii), the equation of one line is
$
y-2=3(x-3) \quad \text { or } \quad 3 x-y=7
$

Taking the negative sign:, $2 m-1=-m-2$
or $\quad 3 m=-1 \quad$ or $\quad m=-\frac{1}{3}$
Putting this value of $m$ in (ii), the equation of second line is
$
y-2=-\frac{1}{3}(x-3) \quad \text { or } \quad 3 y-6=-x+3
$
or $x+3 y=9$
Hence, the equations of two required lines are
$
3 x-y=7, x+3 y=9
$

Remark: Whenever you are asked to find equations of lines (i.e., plural form), start thinking and doing to find the equation of a line (singular form), you will get the required lines.

Q12. Find the equation of the line passing through the point of intersection of the lines $4 x+7 y-3=0$ and $2 x-3 y+1=0$ that has equal intercepts on the axes.

Sol. Given lines are $\quad 4 x+7 y-3=0$
and $2 x-3 y+1=0$

To find point of intersection of lines (i) and (ii), solve them for $x$ and $y$.
$\operatorname{eqn}(i) \times 3+\operatorname{eqn}(i i) \times 7$ gives to eliminate $y$, $12 x+21 y-9+14 x-21 y+7=0$
$\Rightarrow 26 x-2=0 \Rightarrow 26 x=2 \Rightarrow x=\frac{2}{26}=\frac{1}{13}$
Putting $x=\frac{1}{13}$ in (i), $\frac{4}{13}+7 y-3=0$
$\Rightarrow 7 y=3-\frac{4}{13}=\frac{39-4}{13}=\frac{35}{13}$
$\Rightarrow y=\frac{35}{7 \times 13}=\frac{5}{13}$
∴ The point of intersection of given lines is $\mathrm{P}\left(\frac{1}{13}, \frac{5}{13}\right)$.
Required line has equal intercepts on the axes.
Let each intercept be $a$, the equation of line is
$
\frac{x}{a}+\frac{y}{a}=1 \quad \Rightarrow \frac{x+y}{a}=1
$
or
$
x+y=a
$

It passes through $\mathrm{P}\left(\frac{1}{13}, \frac{5}{13}\right)$.
[Therefore, putting $x=\frac{1}{13}, y=\frac{5}{13}$ in (iii),
We have $\frac{1}{13}+\frac{5}{13}=a \quad$ or $\quad a=\overline{13}$

Putting $a=\frac{6}{13}$ in (iii), the required line is
$
x+y=\frac{6}{13} \quad \text { or } \quad 13 x+13 y=6
$

Q13. Show that the equation of the line passing through the origin and making an angle $\theta$ with the line $y=m x+c$ is $ \frac{y}{x}=\frac{m \pm \tan \theta}{1 \mp m \tan \theta} . $

Sol. Given line is
$
l: y=m x+c(\text { slope }- \text { intercept form })
$

Its slope $=m$.
Equation of any line through the origin $\mathrm{O}(0,0)$ is
$
\text { or } \begin{array}{rlr}
y-0 & =\mathrm{M}(x-0) \Rightarrow y=\mathrm{M} x \\
\frac{y}{x} & =\mathrm{M}
\end{array}
$
where M is the slope of the line.

Since the angle between (i) and (ii) is $\theta$, we have
$
\tan \theta=\left|\frac{\mathrm{M}-m}{1+\mathrm{M} m}\right|
$
or
$
\tan \theta= \pm \frac{\mathrm{M}-m}{1+\mathrm{M} m}
$

Taking the positive sign:
$
\begin{aligned}
& \tan \theta=\frac{\mathrm{M}-m}{1+\mathrm{M} m} \\
& \text { Cross-multiplying } \tan \theta(1+\mathrm{M} m)=\mathrm{M}-m \\
& \Rightarrow \quad \tan \theta+\mathrm{M} m \tan \theta=\mathrm{M}-m \Rightarrow \mathrm{~m}+\tan \theta=\mathrm{M}-\mathrm{Mm} \tan \theta \\
& \Rightarrow \quad m+\tan \theta=\mathrm{M}(1-m \tan \theta) \\
& \Rightarrow \quad \mathrm{M}=\frac{m+\tan \theta}{1-m \tan \theta}
\end{aligned}
$
$
\begin{aligned}
& & \tan \theta & =-\frac{\mathrm{M}-m}{1+\mathrm{M} m} \\
& \Rightarrow & \tan \theta(1+\mathrm{M} m) & =-\mathrm{M}+m \\
& \Rightarrow & \tan \theta+\mathrm{M} m \tan \theta= & \mathrm{M}+m \Rightarrow \mathrm{M}+\mathrm{M} m \tan \theta=m-\tan \theta \\
& \Rightarrow & \mathrm{M}(1+m \tan \theta) & =m-\tan \theta \\
& \Rightarrow & \mathrm{M} & =\frac{m-\tan \theta}{1+m \tan \theta}
\end{aligned}
$

Combining the two values of M , we have
$
\mathrm{M}=\frac{m \pm \tan \theta}{1 \mp m \tan \theta}
$

Putting this value of M in (ii), the required equation of line is
$
\frac{y}{x}=\frac{m \pm \tan \theta}{1 \mp m \tan \theta} .
$

Q14. In what ratio, the line joining $(-1,1)$ and $(5,7)$ is divided by the line $x+y=4$ ?

Sol. Let line $l: x+y=4$ meet segment PQ at point R, dividing it in ratio $k:1$. By the section formula:
$
\mathrm{R}=\left(\frac{5 k-1}{k+1}, \frac{7 k+1}{k+1}\right)
$
(By Section Formula)

Since R lies on $l$, its coordinates satisfy (i):
$
\begin{aligned}
& \Rightarrow \quad \frac{5 k-1}{k+1}+\frac{7 k+1}{k+1}=4 \Rightarrow \frac{5 k-1+7 k+1}{k+1}=4 \Rightarrow \frac{12 k}{k+1}=4 \\
& \Rightarrow \quad 12 k=4 k+4 \\
& \Rightarrow \quad 8 k=4 \quad \Rightarrow \quad k=\frac{1}{2}
\end{aligned}
$

Therefore, the line $l$ divides PQ in the ratio $\frac{1}{2}:1$, i.e., $1:2$.

Q15. Find the distance of the line $4 x+7 y+5=0$ from the point ( 1,2 ) along the line $2 x-y=0$.

Sol. Note that $\mathrm{P}(1,2)$ lies on $2x-y=0$ (since $2\times1-2=0$). We find Q, the intersection of
$
\begin{array}{r}
4 x+7 y+5=0 \\
\text { and } \quad 2 x-y=0
\end{array}
$

Solving (i) and (ii) simultaneously to find Q:

From (ii) $2 x=y$, i.e., $y=2 x$
Putting $y=2 x$ in (i), $4 x+14 x+5=0 \Rightarrow 18 x=-5$
$
\begin{aligned}
& \Rightarrow x=-\frac{5}{18} \text { and therefore } y=2 x=-\frac{10}{18}=-\frac{5}{9} \\
& \begin{aligned}
& \therefore \text { Point } \mathrm{Q}=\left(-\frac{5}{18},-\frac{5}{9}\right) . \\
& \therefore \text { Required distance }=\mathrm{PQ}=\sqrt{\left(-\frac{5}{18}-1\right)^2+\left(-\frac{5}{9}-2\right)^2} \\
& \quad=\sqrt{\left(\frac{-5-18}{18}\right)^2+\left(\frac{-5-18}{9}\right)^2}=\sqrt{\left(\frac{-23}{18}\right)^2+\left(\frac{-23}{9}\right)^2} \\
& \quad=\sqrt{\frac{529}{324}+\frac{529}{81}}=\sqrt{\frac{529}{81}\left(\frac{1}{4}+1\right)} \\
& \quad=\sqrt{\frac{529}{81} \times \frac{5}{4}}=\frac{23}{9} \times \frac{\sqrt{5}}{2}=\frac{23 \sqrt{5}}{18}
\end{aligned} \text { units. }
\end{aligned}
$

Q16. Find the direction in which a straight line must be drawn through the point $(-1,2)$ so that its point of intersection with the line $x+y=4$ may be at a distance of 3 units from this point.

Sol. Let the line through $\mathrm{P}(-1,2)$ intersect $x+y=4$ at $(h,k)$. Then $(h,k)$ satisfies:
$
\therefore \quad h+k=4
$

The distance from $(-1,2)$ to $(h,k)$ equals 3:
$
\therefore \quad \sqrt{(h+1)^2+(k-2)^2}=3
$

Putting $k=4-h$ from ( $i$ ),
$
\begin{aligned}
& \sqrt{(h+1)^2+(4-h-2)^2}=3 \\
& (h+1)^2+(2-h)^2=9 \\
& \Rightarrow h^2+1+2 h+4+h^2-4 h=9 \\
& \Rightarrow 2 h^2-2 h-4=0
\end{aligned}
$

⇒ Dividing by $2, h^2-h-2=0 \Rightarrow(h-2)(h+1)=0$
$\Rightarrow \quad h=2$ or -1
When $h=2, k=4-h=4-2=2$
When $h=-1, k=4-h=4-(-1)=5$
∴ The points of intersection of lines through $\mathrm{P}(-1,2)$ with the line $x+y=4$ at a distance of 3 units from P are $\mathrm{Q}(2,2)$ and $\mathrm{R}(-1,5)$.

Slope of $\mathrm{PQ}=\frac{2-2}{2-(-1)}=0$
$\Rightarrow \mathrm{PQ}$ is parallel to $x$-axis.
Slope of $\mathrm{PR}=\frac{5-2}{-1-(-1)}=\frac{3}{0}=\infty$ which is not defined.
⇒ PR is parallel to $y$-axis.
Hence, the required line is parallel to $x$-axis or parallel to $y$-axis.

Q17. The hypotenuse of a right-angled triangle has its ends at the points $(1,3)$ and ( $-4,1$ ). Find the equations of the legs (perpendicular sides) of the triangle.

Sol. Let the hypotenuse endpoints be $\mathrm{A}(1,3)$ and $\mathrm{B}(-4,1)$, and $\mathrm{C}(h,k)$ be the right-angle vertex ($\angle\mathrm{ACB}=90^{\circ}$).

Slope of leg $\mathrm{AC}=\frac{k-3}{h-1}$

Slope of leg $\mathrm{BC}=\frac{k-1}{h+4}$

Since $\mathrm{AC} \perp \mathrm{BC},\left(\because<\mathrm{ACB}=90^{\circ}\right)$
$
\begin{array}{rlrl}
& \therefore\left(\frac{k-3}{h-1}\right)\left(\frac{k-1}{h+4}\right)=-1 & \\
{\left[\because \text { For perpendicularity } m_1 m_2=-1\right]} & \\
(k-3)(k-1) & =-(h-1)(h+4) \\
\Rightarrow \quad(h-1)(h+4)+(k-3)(k-1) & =0
\end{array}
$

Equation (iii) is satisfied by infinitely many pairs $(h,k)$.
Putting $h=1$ and $k=1$ in (iii), $(1-1)(1+4)+(1-3)(1-1)=0$ i.e., $0+0=0$ which is true.
$\therefore h=1$ and $k=1$ satisfy (iii)
∴ co-ordinates of vertex C are $(h, k)=(1,1)$
Putting $h=1, k=1$ in (i), slope of leg AC = $\frac{1-3}{1-1}=-\frac{2}{0} =-\infty$, undefined.

∴ Leg AC is parallel to $y$-axis.
Hence equation of leg AC through $\mathrm{A}(1,3)=\left(x_1, y_1\right)$ is $\boldsymbol{x}=\boldsymbol{x}_{\mathbf{1}}$ i.e., $x=1$
Putting $h=1, k=1$ in (ii), slope of leg $\mathrm{BC}=\frac{1-1}{1+4}=0$
∴ Leg BC is horizontal.
Hence equation of leg BC through $\mathrm{B}(-4,1)=\left(x_1, y_1\right)$ is i.e., $y=1$
Remark. In fact $C$ is any point on the circle drawn on $A B$ as diameter. $C$ is not unique. In any position, the coordinates of C must satisfy (iii) and hence legs AC and BC have infinite positions and hence infinite equations.

Q18. Find the image of the point $(3,8)$ with respect to the line $x+3 y=7$ assuming the line to be a plane mirror.

Sol. The given mirror line is $\mathrm{AB}: x+3y-7=0$.

The given point is $\mathrm{P}(3,8)$. Draw $\mathrm{PM}\perp\mathrm{AB}$; the image Q satisfies $\mathrm{MQ}=\mathrm{PM}$ (the image is as far behind the mirror as the object is in front).

Slope of line AB is $-\frac{a}{b}=\frac{-1}{3}$.
So the slope of PM (perpendicular to AB) is 3 (negative reciprocal).
Using the point-slope form, the equation of PM is
$
\begin{aligned}
& y-8=3(x-3) \text { or } y-8=3 x-9 \\
& \text { or }-3 x+y+1=0 \text { or } 3 x-y-1=0
\end{aligned}
$

To find foot M, we solve lines (i) and (ii) simultaneously.
$\operatorname{Eqn}(i)+3 \times \operatorname{Eqn}(i i)$ gives
$x+3 y-7+9 x-3 y-3=0$ or $10 x-10=0$
or $10 x=10 \Rightarrow x=\frac{10}{10}=1$
Putting $x=1$ in ( $i$ ), $1+3 y-7=0$ or $3 y=6$ i.e., $y=2$
∴ Foot of perpendicular M is $(1,2)$

Let the image point be $\mathrm{Q}(\alpha,\beta)$. Since M is the midpoint of PQ:
$
\begin{array}{rlrl}
\therefore & \frac{\alpha+3}{2} & =1 \quad \text { and } \quad \frac{\beta+8}{2} & =2 \quad \therefore \alpha+3=2 \text { and } \beta+8=4 . \\
\Rightarrow & \alpha & =-1 \text { and } & \beta \\
& =-4 .
\end{array}
$

Hence, the image of P in line AB is $\mathrm{Q}(-1,-4)$.

Q19. If the lines $y=3 x+1$ and $2 y=x+3$ are equally inclined to the line $y=m x+4$, find the value of $m$.

Sol. The given lines are
$
\begin{aligned}
y & =3 x+1 \\
2 y & =x+3 \text { or } y=\frac{1}{2} x+\frac{3}{2} \\
y & =m x+c
\end{aligned}
$

All three equations are in slope-intercept form, so their slopes are the coefficients of $x$:
Their slopes are $3$, $\frac{1}{2}$, and $m$ respectively. Since (i) and (ii) make equal angles with (iii):
Then, $\tan \theta=\left|\frac{m-3}{1+m \cdot 3}\right|=\left|\frac{m-\frac{1}{2}}{1+m \cdot \frac{1}{2}}\right|$
$
\begin{gathered}
\Rightarrow \quad\left|\frac{m-3}{1+3 m}\right|=\left|\frac{2 m-1}{2+m}\right| \Rightarrow \frac{m-3}{1+3 m}= \pm \frac{2 m-1}{2+m} \\
{[\because|x|=|y| \Rightarrow x= \pm y]} \\
\operatorname{Cross-multiplying}(m-3)(2+\mathrm{m})= \pm(1+3 m)(2 m-1)
\end{gathered}
$

Taking the positive sign:
$
\begin{aligned}
& & (m-3)(2+m) & =(1+3 m)(2 m-1) \\
\Rightarrow & & m^2-m-6 & =6 m^2-m-1 \\
\Rightarrow & & -5 m^2 & =5 \\
\Rightarrow & & m^2 & =-1<0 \text { which is not possible. }
\end{aligned}
$

Taking the negative sign:
$
\begin{array}{rlrl}
& & (m-3)(2+m) & =-(1+3 m)(2 m-1) \\
\Rightarrow & & m^2-m-6=-6 m^2+m+1 \\
\Rightarrow & & 7 m^2-2 m-7=0 \\
& \Rightarrow & m=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a} \quad=\frac{2 \pm \sqrt{4+196}}{14}=\frac{2 \pm \sqrt{200}}{14} \\
& & =\frac{2 \pm 10 \sqrt{2}}{14}=\frac{1 \pm 5 \sqrt{2}}{7} .
\end{array}
$

Q20. If sum of the perpendicular distances of a variable point $\mathbf{P}(x, y)$ from the lines $x+y-5=0$ and $3 x-2 y+7=0$ is always 10 , show that $P$ must move on a line.

Sol. The two given lines are $x+y-5=0$ and $3x-2y+7=0$. Since the sum of perpendicular distances from $\mathrm{P}(x,y)$ to both lines equals 10:
$
\begin{array}{ll}
\therefore & \frac{|x+y-5|}{\sqrt{1^2+1^2}}+\frac{|3 x-2 y+7|}{\sqrt{3^2+(-2)^2}}=10 \\
\Rightarrow & \pm \frac{x+y-5}{\sqrt{2}} \pm \frac{3 x-2 y+7}{\sqrt{13}}=10 \\
\Rightarrow & \pm \sqrt{13}(x+y-5) \pm \sqrt{2}(3 x-2 y+7)=10 \sqrt{26}
\end{array}
$

Rearranging these equations yields a first-degree equation in $x$ and $y$ of the form $Ax+By+C=0$, which represents a straight line.
Therefore, $\mathrm{P}(x,y)$ must move along a straight line.

Q21. Find equation of the line which is equidistant from parallel lines $9 x+6 y-7=0$ and $3 x+2 y+6=0$.

Sol. The parallel lines are $9x+6y-7=0$ and $3x+2y+6=0$. Multiply the second by 3 to match the coefficients of $x$ and $y$ in (i):
$
9 x+6 y+18=0
$

Any line equidistant from (i) and (ii) is also parallel to them. Let it be $9x+6y+k=0$.

Setting the distance from (iii) to (i) equal to the distance from (iii) to (ii):
$
\begin{aligned}
\Rightarrow & \frac{|k-(-7)|}{\sqrt{9^2+2^2}} & =\frac{|k-18|}{\sqrt{9^2+2^2}} & \quad\left[d=\frac{\left|1_2\right|}{\sqrt{a^2+b^2}}\right. \\
\Rightarrow & |k+7| & =|k-18| & \\
\Rightarrow & k+7 & = \pm(k-18) & {[|x|=|y| \Rightarrow x= \pm y] }
\end{aligned}
$

Taking the positive sign:
$
k+7=k-18
$
This gives $7=-18$, which is impossible.
Taking the negative sign:
$
k+7=-(k-18)
$

This gives $2k=11$, so $k=\frac{11}{2}$. Substituting into (iii), the required line is $18x+12y+11=0$.

Q22. A ray of light passing through the point ( 1,2 ) reflects on the $\boldsymbol{x}$-axis at point $\mathbf{A}$ and the reflected ray passes through the point $(5,3)$. Find the coordinates of $\mathbf{A}$.

Sol. Let $\mathrm{A}(\alpha,0)$ be on the $x$-axis. By the Law of Reflection, the incident ray PA (from $P(1,2)$) and reflected ray AQ (toward $Q(5,3)$) make equal angles $\theta$ with the normal AN.

With $\angle\mathrm{XAQ}=90^{\circ}-\theta$ and $\angle\mathrm{XAP}=90^{\circ}+\theta$: slope of AQ $=\cot\theta$ and slope of AP $=-\cot\theta$. Their sum is zero:
$
\begin{aligned}
\Rightarrow & & \frac{3-0}{5-a}+\frac{2-0}{1-a} & =0 \Rightarrow \\
\Rightarrow & & 3-3 a+10-2 a & =0 \Rightarrow \\
\therefore & & a & =\frac{13}{5}
\end{aligned}
$

Therefore, the point of reflection is $\mathrm{A}\left(\frac{13}{5}, 0\right)$.

Q23. Prove that the product of the lengths of the perpendiculars drawn from the points $\left(\sqrt{a^2-b^2}, 0\right)$ and $\left(-\sqrt{a^2-b^2}, 0\right)$ to the line $\frac{x}{a} \cos \theta+\frac{y}{b} \sin \theta=1$ is $b^2$.

Sol. Write the given line as $\frac{x}{a}\cos\theta+\frac{y}{b}\sin\theta-1=0$.
Compute $p_1$, the perpendicular distance from $\left(\sqrt{a^2-b^2}, 0\right)$ to line (i):
$
\begin{aligned}
& \text { using the formula } \left.\frac{\left|a x_1+b y_1+c\right|}{\sqrt{a^2+b^2}}\right) \\
& =\frac{\left|\frac{\sqrt{a^2-b^2}}{a} \cos \theta+0-1\right|}{\sqrt{\left(\frac{\cos \theta}{a}\right)^2+\left(\frac{\sin \theta}{b}\right)^2}}=\frac{\left|\frac{\sqrt{a^2-b^2}}{a} \cos \theta-1\right|}{\sqrt{\frac{\cos ^2 \theta}{a^2}+\frac{\sin ^2 \theta}{b^2}}}
\end{aligned}
$
Compute $p_2$, the perpendicular distance from $\left(-\sqrt{a^2-b^2}, 0\right)$ to line (i):
$
\begin{aligned}
& =\frac{\left|\frac{-\sqrt{a^2-b^2}}{a} \cos \theta+0-1\right|}{\sqrt{\left(\frac{\cos \theta}{a}\right)^2+\left(\frac{\sin \theta}{b}\right)^2}}=\frac{\left|-\left(\frac{\sqrt{a^2-b^2}}{a} \cos \theta+1\right)\right|}{\sqrt{\frac{\cos ^2 \theta}{a^2}+\frac{\sin ^2 \theta}{b^2}}} \\
& =\frac{\left|\frac{\sqrt{a^2-b^2}}{a} \cos \theta+1\right|}{\sqrt{\frac{\cos ^2 \theta}{a^2}+\frac{\sin ^2 \theta}{b^2}}}[\because|-x|=|x|] \\
& \left.\therefore p_1 p_2=\frac{\mid \sqrt{a^2-b^2}}{a} \cos \theta-1 \right\rvert\, \\
& \sqrt{\frac{\cos ^2 \theta}{a^2}+\frac{\sin ^2 \theta}{b^2}} \times \frac{\left|\frac{\sqrt{a^2-b^2}}{a} \cos \theta+1\right|}{\sqrt{\frac{\cos ^2 \theta}{a^2}+\frac{\sin ^2 \theta}{b^2}}} \\
& =\frac{\left|\left(\frac{\sqrt{a^2-b^2}}{a} \cos \theta-1\right)\left(\frac{\sqrt{a^2-b^2}}{a} \cos \theta+1\right)\right|}{\frac{\cos ^2 \theta}{a^2}+\frac{\sin ^2 \theta}{b^2}} \quad[\because|x||y|=|x y|]
\end{aligned}
$

$
\begin{aligned}
& =\frac{\left|\frac{a^2-b^2}{a^2} \cos ^2 \theta-1\right|}{\frac{\cos ^2 \theta}{a^2}+\frac{\sin ^2 \theta}{b^2}}=\frac{\left|\frac{\left(a^2-b^2\right) \cos ^2 \theta-a^2}{a^2}\right|}{\frac{b^2 \cos ^2 \theta+a^2 \sin ^2 \theta}{a^2 b^2}} \\
& =\frac{\left|a^2 \cos ^2 \theta-b^2 \cos ^2 \theta-a^2\right|}{a^2} \times \frac{a^2 b^2}{b^2 \cos ^2 \theta+a^2 \sin ^2 \theta} \\
& =\frac{b^2\left|-a^2\left(1-\cos ^2 \theta\right)-b^2 \cos ^2 \theta\right|}{b^2 \cos ^2 \theta+a^2 \sin ^2 \theta} \\
& =\frac{b^2\left|-a^2 \sin ^2 \theta-b^2 \cos ^2 \theta\right|}{b^2 \cos ^2 \theta+a^2 \sin ^2 \theta}=\frac{b^2\left|-\left(b^2 \cos ^2 \theta+a^2 \sin ^2 \theta\right)\right|}{b^2 \cos ^2 \theta+a^2 \sin ^2 \theta} \\
& =\frac{b^2\left|b^2 \cos ^2 \theta+a^2 \sin ^2 \theta\right|}{b^2 \cos ^2 \theta+a^2 \sin ^2 \theta} \quad[\because \quad|-x|=|x|] \\
& =\frac{b^2\left(b^2 \cos ^2 \theta+a^2 \sin ^2 \theta\right)}{b^2 \cos ^2 \theta+a^2 \sin ^2 \theta}=b^2 . \quad[\because \quad|x|=x \text { if } x>0]
\end{aligned}
$

This proves $p_1 p_2 = b^2$.

Q24. A person standing at the junction (crossing) of two straight paths represented by the equations $2 x-3 y+4=0$ and $3 x+4 y-5=0$ wants to reach the path whose equation is $6 x-7 y+8=0$ in the least time. Find equation of the path that he should follow.

Sol. The equations of the two paths are
$
\text { and } \quad \begin{aligned}
& 2 x-3 y+4=0 \\
& 3 x+4 y-5=0
\end{aligned}
$

Let us solve eqn (i) and (ii) for $x$ and $y$ to find co-ordinates of the point P [Junction or crossing of paths (i) and (ii)] Eqn (i) $\times 4+$ Eqn (ii) $\times 3$ gives to eliminate y.
$
8 x-12 y+16+9 x+12 y-15=0
$
or $17 x+1=0$ or $17 x=-1 \Rightarrow x=\frac{-1}{17}$
Putting $x=\frac{-1}{17}$ in $(i),-\frac{2}{17}-3 y+4=0$
$
\begin{aligned}
& \Rightarrow-3 y=-4+\frac{2}{17}=\frac{-68+2}{17}=\frac{-66}{17} \\
& \Rightarrow y=\frac{66}{3 \times 17}=\frac{22}{17}
\end{aligned}
$
So the junction is at $\mathrm{P}\left(-\frac{1}{17}, \frac{22}{17}\right)$.
The target path AB has equation
$
6 x-7 y+8=0
$

For the shortest time, the person must walk along PQ where $\mathrm{PQ}\perp\mathrm{AB}$ (since perpendicular distance is the shortest).
From (iii), slope of $\mathrm{AB}=-\frac{a}{b}=-\frac{6}{-7}=\frac{6}{7}$.
Since $\mathrm{PQ}\perp\mathrm{AB}$, slope of PQ $=-\frac{7}{6}$ (negative reciprocal of $\frac{6}{7}$).
The equation of PQ through $\mathrm{P}\left(-\frac{1}{17}, \frac{22}{17}\right)$ with slope $-\frac{7}{6}$ is
$
\begin{aligned}
& & y-\frac{22}{17} & =-\frac{7}{6}\left(x+\frac{1}{17}\right) \\
& \Rightarrow & 6 y-\frac{132}{17} & =-7 x-\frac{7}{17} \\
& \Rightarrow & 7 x+6 y & =\frac{125}{17}
\end{aligned}
$
[Point-slope Form]
Cross-multiplying gives $119x+102y=125$.