Class 10 NCERT Solutions
Chapter 7: Coordinate Geometry
Master the distance formula, the section formula, and the logic of the Cartesian plane to precisely locate and divide points with our step-by-step visual guidance.
Exercise 7.1
1. Find the distance between the following pairs of points:
(i) $(2,3),(4,1)$
(ii) $(-5,7),(-1,3)$
(iii) $(a,b),(-a,-b)$
Apply the distance formula $d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ to each pair.
(i)
$$\text{Distance} = \sqrt{(4-2)^2+(1-3)^2}=\sqrt{4+4}=\sqrt{8}=2\sqrt{2} \text{ units.}$$(ii)
$$\text{Distance} = \sqrt{(-1+5)^2+(3-7)^2}=\sqrt{16+16}=4\sqrt{2} \text{ units.}$$(iii) Let points be $\text{A}(a,b)$ and $\text{B}(-a,-b)$, then
$$\text{AB} = \sqrt{(-a-a)^2+(-b-b)^2}=\sqrt{(-2a)^2+(-2b)^2}=\sqrt{4a^2+4b^2}=2\sqrt{a^2+b^2} \text{ units.}$$
2. Find the distance between the points $(0,0)$ and $(36,15)$. Can you now find the distance between the two towns A and B: “A town B is located 36 km east and 15 km north of the town A”.
Yes, we can find the distance between the two towns as follows.
From the figure, the two towns are situated at $\text{A}(0,0)$ and $\text{B}(36,15)$.
$\therefore$ Distance, $\text{AB} = 39$ km.
3. Determine if the points $(1,5),(2,3)$ and $(-2,-11)$ are collinear.
Let points be $\text{A}(1,5)$, $\text{B}(2,3)$ and $\text{C}(-2,-11)$. We compute all three pairwise distances.
$$\text{AB} = \sqrt{(2-1)^2+(3-5)^2}=\sqrt{1+4}=\sqrt{5}$$ $$\text{BC} = \sqrt{(-2-2)^2+(-11-3)^2}=\sqrt{16+196}=\sqrt{212}=2\sqrt{53}$$ $$\text{AC} = \sqrt{(-2-1)^2+(-11-5)^2}=\sqrt{9+256}=\sqrt{265}$$Since none of these distances satisfies the condition that one equals the sum of the other two, the three points are not collinear.
4. Check whether $(5,-2),(6,4)$ and $(7,-2)$ are the vertices of an isosceles triangle.
Let points be $\text{A}(5,-2)$, $\text{B}(6,4)$ and $\text{C}(7,-2)$. We find all three side lengths.
$$\text{AB} = \sqrt{(6-5)^2+(4+2)^2}=\sqrt{1+36}=\sqrt{37}$$ $$\text{BC} = \sqrt{(7-6)^2+(-2-4)^2}=\sqrt{1+36}=\sqrt{37}$$ $$\text{AC} = \sqrt{(7-5)^2+(-2+2)^2}=\sqrt{4+0}=2$$Since $\text{AB} = \text{BC}$ (two sides are equal) and $\text{AC}$ is different, and no distance equals the sum of the other two, the given points are indeed the vertices of an isosceles triangle.
5. In a classroom, 4 friends are seated at the points A, B, C and D as shown in figure. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, “Don’t you think ABCD is a square?” Chameli disagrees. Using distance formula, find which of them is correct.
From the figure, the coordinates are $\text{A}(3,4)$, $\text{B}(6,7)$, $\text{C}(9,4)$ and $\text{D}(6,1)$. We check all four sides and both diagonals.
$$\text{AB} = \sqrt{(6-3)^2+(7-4)^2}=\sqrt{9+9}=3\sqrt{2}$$ $$\text{BC} = \sqrt{(9-6)^2+(4-7)^2}=\sqrt{9+9}=3\sqrt{2}$$ $$\text{CD} = \sqrt{(6-9)^2+(1-4)^2}=\sqrt{9+9}=3\sqrt{2}$$ $$\text{DA} = \sqrt{(6-3)^2+(1-4)^2}=\sqrt{9+9}=3\sqrt{2}$$ $$\text{AC} = \sqrt{(9-3)^2+(4-4)^2}=\sqrt{36+0}=6$$ $$\text{BD} = \sqrt{(6-6)^2+(1-7)^2}=\sqrt{0+36}=6$$All four sides are equal and both diagonals are equal. Therefore, ABCD is a square, and Champa is correct.
6. Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:
(i) $(-1,-2),(1,0),(-1,2),(-3,0)$
(ii) $(-3,5),(3,1),(0,3),(-1,-4)$
(iii) $(4,5),(7,6),(4,3),(1,2)$
(i) Let points be $\text{A}(-1,-2)$, $\text{B}(1,0)$, $\text{C}(-1,2)$ and $\text{D}(-3,0)$.
$$\text{AB} = \sqrt{(1+1)^2+(0+2)^2}=\sqrt{4+4}=2\sqrt{2}$$ $$\text{BC} = \sqrt{(-1-1)^2+(2-0)^2}=\sqrt{4+4}=2\sqrt{2}$$ $$\text{CD} = \sqrt{(-3+1)^2+(0-2)^2}=\sqrt{4+4}=2\sqrt{2}$$ $$\text{DA} = \sqrt{(-1+3)^2+(-2-0)^2}=\sqrt{4+4}=2\sqrt{2}$$ $$\text{AC} = \sqrt{(-1+1)^2+(2+2)^2}=\sqrt{0+16}=4$$ $$\text{BD} = \sqrt{(-3-1)^2+(0-0)^2}=\sqrt{16+0}=4$$All four sides $\text{AB}$, $\text{BC}$, $\text{CD}$, $\text{DA}$ are equal and diagonals $\text{AC}$, $\text{BD}$ are also equal. Hence, quadrilateral ABCD is a square.
(ii) Let points be $\text{A}(-3,5)$, $\text{B}(3,1)$, $\text{C}(0,3)$ and $\text{D}(-1,-4)$.
$$\text{AB} = \sqrt{(3+3)^2+(1-5)^2}=\sqrt{36+16}=\sqrt{52}=2\sqrt{13}$$ $$\text{BC} = \sqrt{(0-3)^2+(3-1)^2}=\sqrt{9+4}=\sqrt{13}$$ $$\text{AC} = \sqrt{(0+3)^2+(3-5)^2}=\sqrt{9+4}=\sqrt{13}$$Since $\text{AB} = \text{AC} + \text{BC}$, i.e., $2\sqrt{13} = \sqrt{13}+\sqrt{13}$, the points A, B and C are collinear. Hence, no quadrilateral exists.
(iii) Let points be $\text{A}(4,5)$, $\text{B}(7,6)$, $\text{C}(4,3)$ and $\text{D}(1,2)$.
$$\text{AB} = \sqrt{(7-4)^2+(6-5)^2}=\sqrt{9+1}=\sqrt{10}$$ $$\text{BC} = \sqrt{(4-7)^2+(3-6)^2}=\sqrt{9+9}=3\sqrt{2}$$ $$\text{CD} = \sqrt{(1-4)^2+(2-3)^2}=\sqrt{9+1}=\sqrt{10}$$ $$\text{DA} = \sqrt{(4-1)^2+(5-2)^2}=\sqrt{9+9}=3\sqrt{2}$$ $$\text{AC} = \sqrt{(4-4)^2+(3-5)^2}=\sqrt{0+4}=2$$ $$\text{BD} = \sqrt{(1-7)^2+(2-6)^2}=\sqrt{36+16}=2\sqrt{13}$$Since $\text{AB} = \text{CD}$, $\text{BC} = \text{DA}$, but $\text{AC} \neq \text{BD}$, the quadrilateral is a parallelogram.
7. Find the point on the $x$-axis which is equidistant from $(2,-5)$ and $(-2,9)$.
Let $\text{P}(x,0)$ on the $x$-axis be equidistant from $\text{A}(2,-5)$ and $\text{B}(-2,9)$.
Setting $\text{AP} = \text{BP}$:
$$\sqrt{(x-2)^2+(0+5)^2} = \sqrt{(x+2)^2+(0-9)^2}$$ $$\Rightarrow x^2-4x+4+25 = x^2+4x+4+81$$ $$\Rightarrow -8x = 56 \quad \Rightarrow x = -7$$$\therefore$ The required point is $(-7,0)$.
8. Find the values of $y$ for which the distance between the points $P(2,-3)$ and $Q(10,y)$ is 10 units.
Setting the distance equal to 10 and squaring both sides:
$$\sqrt{(10-2)^2+(y+3)^2} = 10 \Rightarrow 64+9+6y+y^2 = 100$$ $$\Rightarrow y^2+6y-27 = 0 \Rightarrow (y+9)(y-3) = 0$$ $$\Rightarrow y = -9,\ 3$$
9. If $Q(0,1)$ is equidistant from $P(5,-3)$ and $R(x,6)$, find the values of $x$. Also find the distances $QR$ and $PR$.
Since $\text{Q}(0,1)$ is equidistant from $\text{P}(5,-3)$ and $\text{R}(x,6)$, we set $\text{PQ} = \text{RQ}$:
$$\sqrt{(0-5)^2+(1+3)^2} = \sqrt{(0-x)^2+(1-6)^2}$$Squaring and simplifying:
$$25+16 = x^2+25 \Rightarrow x^2 = 16 \Rightarrow x = \pm 4$$So point R is either $\text{R}(4,6)$ or $\text{R}(-4,6)$.
With $\text{R}(4,6)$:
$$\text{QR} = \sqrt{(4-0)^2+(6-1)^2}=\sqrt{16+25}=\sqrt{41}$$ $$\text{PR} = \sqrt{(4-5)^2+(6+3)^2}=\sqrt{1+81}=\sqrt{82}$$With $\text{R}(-4,6)$:
$$\text{QR} = \sqrt{(-4-0)^2+(6-1)^2}=\sqrt{16+25}=\sqrt{41}$$ $$\text{PR} = \sqrt{(-4-5)^2+(6+3)^2}=\sqrt{81+81}=9\sqrt{2}$$
10. Find a relation between $x$ and $y$ such that the point $(x,y)$ is equidistant from the points $(3,6)$ and $(-3,4)$.
Let $\text{P}(x,y)$ be equidistant from $\text{A}(3,6)$ and $\text{B}(-3,4)$. Setting $\text{AP} = \text{BP}$ and squaring:
$$\sqrt{(x-3)^2+(y-6)^2} = \sqrt{(x+3)^2+(y-4)^2}$$ $$\Rightarrow x^2-6x+9+y^2-12y+36 = x^2+6x+9+y^2-8y+16$$ $$\Rightarrow -12x-4y+20 = 0$$ $$\Rightarrow 3x+y-5 = 0 \text{ is the required relation.}$$Exercise 7.2
1. Find the coordinates of the point which divides the join of $(-1,7)$ and $(4,-3)$ in the ratio $2:3$.
Using the section formula directly:
Coordinates of point $= \left(\dfrac{2 \times 4 + 3 \times (-1)}{2+3},\ \dfrac{2 \times (-3) + 3 \times 7}{2+3}\right) = \left(\dfrac{8-3}{5},\ \dfrac{-6+21}{5}\right) = (1,\ 3)$.
2. Find the coordinates of the points of trisection of the line segment joining $(4,-1)$ and $(-2,-3)$.
Let P and Q be the points of trisection of segment AB where $\text{A}(4,-1)$ and $\text{B}(-2,-3)$.
P divides AB in the ratio $1:2$, so:
$$\text{P} = \left(\frac{-2+8}{1+2},\ \frac{-3-2}{1+2}\right) = \text{P}\!\left(2,\ -\frac{5}{3}\right)$$Q divides AB in the ratio $2:1$, so:
$$\text{Q} = \left(\frac{-4+4}{3},\ \frac{-6-1}{3}\right) = \text{Q}\!\left(0,\ -\frac{7}{3}\right)$$
3. To conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1 m each. 100 flower pots have been placed at a distance of 1 m from each other along AD. Niharika runs $\frac{1}{4}$th the distance AD on the 2nd line and posts a green flag. Preet runs $\frac{1}{5}$th the distance AD on the eighth line and posts a red flag. What is the distance between both the flags? If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her flag?
Position of green flag: $\text{G}(2,25)$. Position of red flag: $\text{R}(8,20)$.
Distance between the two flags:
$$\text{GR} = \sqrt{(8-2)^2+(20-25)^2} = \sqrt{36+25} = \sqrt{61} \text{ m}$$Rashmi posts a blue flag at the midpoint of GR:
$$\text{Midpoint} = \left(\frac{2+8}{2},\ \frac{25+20}{2}\right) = (5,\ 22.5)$$Therefore, Rashmi should move to the 5th line at a distance of 22.5 m.
4. Find the ratio in which the line segment joining the points $(-3,10)$ and $(6,-8)$ is divided by $(-1,6)$.
Let the ratio be $k:1$. Using the section formula, the dividing point is:
$$\left(\frac{6k-3}{k+1},\ \frac{-8k+10}{k+1}\right) = (-1,\ 6)$$From the $x$-coordinate: $6k-3 = -(k+1) \Rightarrow 7k = 2 \Rightarrow k = \dfrac{2}{7}$
Verifying with the $y$-coordinate gives the same result. Therefore, the required ratio is $\dfrac{2}{7}:1$, i.e., $\mathbf{2:7}$.
5. Find the ratio in which the line segment joining $A(1,-5)$ and $B(-4,5)$ is divided by the $x$-axis. Also find the coordinates of the point of division.
Let the ratio be $k:1$. The point of division is $\left(\dfrac{-4k+1}{k+1},\ \dfrac{5k-5}{k+1}\right)$.
Since the point lies on the $x$-axis, the $y$-coordinate must be zero:
$$\frac{5k-5}{k+1} = 0 \Rightarrow k = 1$$The ratio is $\mathbf{1:1}$ and the point of division is:
$$\left(\frac{-4+1}{2},\ 0\right) = \left(-\frac{3}{2},\ 0\right)$$
6. If $(1,2),(4,y),(x,6)$ and $(3,5)$ are the vertices of a parallelogram taken in order, find $x$ and $y$.
The diagonals of a parallelogram bisect each other, so their midpoints must be equal:
$$\left(\frac{1+x}{2},\ \frac{2+6}{2}\right) = \left(\frac{3+4}{2},\ \frac{5+y}{2}\right)$$From the $x$-coordinates: $\dfrac{1+x}{2} = \dfrac{7}{2} \Rightarrow x = 6$
From the $y$-coordinates: $4 = \dfrac{5+y}{2} \Rightarrow y = 3$
7. Find the coordinates of a point $A$, where $AB$ is the diameter of a circle whose centre is $(2,-3)$ and $B$ is $(1,4)$.
The centre is the midpoint of the diameter AB, so:
$$\left(\frac{x+1}{2},\ \frac{y+4}{2}\right) = (2,\ -3)$$ $$\Rightarrow \frac{x+1}{2} = 2 \Rightarrow x = 3 \quad \text{and} \quad \frac{y+4}{2} = -3 \Rightarrow y = -10$$Hence, coordinates of A are $(3,-10)$.
8. If $A$ and $B$ are $(-2,-2)$ and $(2,-4)$ respectively, find the coordinates of P such that $AP = \frac{3}{7}AB$ and P lies on the line segment AB.
Since $\text{AP} = \dfrac{3}{7}\text{AB}$:
$$7\text{AP} = 3(\text{AP}+\text{PB}) \Rightarrow 4\text{AP} = 3\text{PB} \Rightarrow \frac{\text{AP}}{\text{PB}} = \frac{3}{4}$$So AP : PB $= 3:4$. Using the section formula:
$$\text{P} = \left(\frac{3 \times 2 + 4 \times (-2)}{3+4},\ \frac{3 \times (-4) + 4 \times (-2)}{3+4}\right) = \left(\frac{6-8}{7},\ \frac{-12-8}{7}\right) = \left(-\frac{2}{7},\ -\frac{20}{7}\right)$$
9. Find the coordinates of the points which divide the line segment joining $A(-2,2)$ and $B(2,8)$ into four equal parts.
Let $\text{P}_1$, $\text{P}_2$ and $\text{P}_3$ divide AB into four equal parts. We find each point by successive midpoints.
$\text{P}_2$ is the midpoint of AB:
$$\text{P}_2 = \left(\frac{-2+2}{2},\ \frac{2+8}{2}\right) = (0,\ 5)$$$\text{P}_1$ is the midpoint of $\text{A}\,\text{P}_2$:
$$\text{P}_1 = \left(\frac{-2+0}{2},\ \frac{2+5}{2}\right) = \left(-1,\ \frac{7}{2}\right)$$$\text{P}_3$ is the midpoint of $\text{P}_2\,\text{B}$:
$$\text{P}_3 = \left(\frac{0+2}{2},\ \frac{5+8}{2}\right) = \left(1,\ \frac{13}{2}\right)$$
10. Find the area of a rhombus if its vertices are $(3,0),(4,5),(-1,4)$ and $(-2,-1)$ taken in order.
[Hint: Area of a rhombus $= \frac{1}{2}$ (product of its diagonals)]
First find the lengths of the diagonals AC and BD:
$$\text{AC} = \sqrt{(-1-3)^2+(4-0)^2}=\sqrt{16+16}=4\sqrt{2}$$ $$\text{BD} = \sqrt{(-2-4)^2+(-1-5)^2}=\sqrt{36+36}=6\sqrt{2}$$ $$\therefore \text{ Area of rhombus} = \frac{1}{2} \times \text{AC} \times \text{BD} = \frac{1}{2} \times 4\sqrt{2} \times 6\sqrt{2} = 24 \text{ sq. units.}$$