CBSE · NCERT · Class 10 Maths · Chapter 7

NCERT Solutions: Class 10 Maths Chapter 7 - Coordinate Geometry

20 textbook Q&A20 verifiedFree Content

Chapter-wise NCERT intext questions and exercise answers for Coordinate Geometry, grounded in the official textbook.

Questions are taken verbatim from the NCERT textbook; answers were grounded against the chapter's content during generation. Items needing review are marked.
Sections in this chapter
Exercise 7.1 10Exercise 7.2 10
Your Progress - Chapter 70% complete
1Exercise 7.110 questions
Q.1Find the distance between the following pairs of points : (i) (2, 3), (4, 1) (ii) (- 5, 7), (- 1, 3) (iii) (a, b), (- a, - b)v
Solution

Use $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$. Thus (i) $d=\sqrt{(4-2)^2+(1-3)^2}=2\sqrt2$, (ii) $d=\sqrt{4^2+(-4)^2}=4\sqrt2$, and (iii) $d=\sqrt{(-2a)^2+(-2b)^2}=2\sqrt{a^2+b^2}$.

Answer:

(i) $2\sqrt{2}$
(ii) $4\sqrt{2}$
(iii) $2\sqrt{a^2+b^2}$

✓ Verified and validated
Q.2Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B discussed in Section 7.2.v
Solution

$d=\sqrt{36^2+15^2}=\sqrt{1296+225}=\sqrt{1521}=39$. Since Section 7.2 uses 1 km as one unit, the distance between the towns is 39 km.

Answer:

The distance is 39 units, so the distance between the towns is 39 km.

✓ Verified and validated
Q.3Determine if the points (1, 5), (2, 3) and (- 2, - 11) are collinear.v
Solution

The slope through $(1,5)$ and $(2,3)$ is $\dfrac{3-5}{2-1}=-2$. The slope through $(2,3)$ and $(-2,-11)$ is $\dfrac{-11-3}{-2-2}=\dfrac{7}{2}$. Since the slopes are not equal, the points are not collinear.

Answer:

The points are not collinear.

✓ Verified and validated
Q.4Check whether (5, - 2), (6, 4) and (7, - 2) are the vertices of an isosceles triangle.v
Solution

Let the points be A$(5,-2)$, B$(6,4)$ and C$(7,-2)$. Then $AB=\sqrt{1^2+6^2}=\sqrt{37}$, $BC=\sqrt{1^2+(-6)^2}=\sqrt{37}$ and $AC=2$. Since $AB=BC$, the triangle is isosceles.

Answer:

Yes, they form an isosceles triangle.

✓ Verified and validated
Q.5In a classroom, 4 friends are seated at the points A, B, C and D as shown in Fig. 7.8. 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. In Fig. 7.8, A(3, 4), B(6, 7), C(9, 4) and D(6, 1).v
Solution

$AB=BC=CD=DA=\sqrt{18}=3\sqrt2$. Also, $AC=6$ and $BD=6$. Since all four sides are equal and the diagonals are equal, ABCD is a square.

Answer:

Champa is correct; ABCD is a square.

✓ Verified and validated
Q.6Name 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)v
Solution

(i) The four sides are all $\sqrt8$ and both diagonals are 4, so it is a square. (ii) The side lengths are $\sqrt{52},\sqrt{13},\sqrt{58},\sqrt{82}$, so there is no standard special quadrilateral from these data. (iii) Opposite sides are equal: $AB=CD=\sqrt{10}$ and $BC=DA=\sqrt{18}$, so the quadrilateral is a parallelogram.

Answer:

(i) A square.
(ii) A quadrilateral with no special type indicated by equal/parallel side tests.
(iii) A parallelogram.

✓ Verified and validated
Q.7Find the point on the x-axis which is equidistant from (2, -5) and (-2, 9).v
Solution

A point on the x-axis is $(x,0)$. Equating squared distances gives $(x-2)^2+5^2=(x+2)^2+9^2$. Hence $x^2-4x+29=x^2+4x+85$, so $x=-7$.

Answer:

The point is $(-7,0)$.

✓ Verified and validated
Q.8Find the values of y for which the distance between the points P(2, - 3) and Q(10, y) is 10 units.v
Solution

$10^2=(10-2)^2+(y+3)^2$, so $100=64+(y+3)^2$. Thus $(y+3)^2=36$, giving $y=3$ or $y=-9$.

Answer:

$y=3$ or $y=-9$.

✓ Verified and validated
Q.9If 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.v
Solution

$QP^2=(5-0)^2+(-3-1)^2=41$. Also $QR^2=x^2+(6-1)^2=x^2+25$. Equating gives $x^2+25=41$, so $x=\pm4$. Then compute $PR$ from P to each possible R.

Answer:

$x=4$ or $x=-4$. In both cases $QR=\sqrt{41}$. If $x=4$, $PR=\sqrt{82}$; if $x=-4$, $PR=9\sqrt2$.

✓ Verified and validated
Q.10Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (- 3, 4).v
Solution

Equating squared distances: $(x-3)^2+(y-6)^2=(x+3)^2+(y-4)^2$. Expanding and simplifying gives $3x+y-5=0$.

Answer:

$3x+y-5=0$.

✓ Verified and validated
2Exercise 7.210 questions
Q.1Find the coordinates of the point which divides the join of (-1, 7) and (4, -3) in the ratio 2 : 3.v
Solution

Using section formula for ratio $2:3$, $x=\dfrac{2(4)+3(-1)}{5}=1$ and $y=\dfrac{2(-3)+3(7)}{5}=3$.

Answer:

The point is $(1,3)$.

✓ Verified and validated
Q.2Find the coordinates of the points of trisection of the line segment joining (4, -1) and (-2, -3).v
Solution

The two points divide the segment in ratios $1:2$ and $2:1$. Applying the section formula to A$(4,-1)$ and B$(-2,-3)$ gives $(2,-\dfrac{5}{3})$ and $(0,-\dfrac{7}{3})$.

Answer:

The points of trisection are $(2,-\dfrac{5}{3})$ and $(0,-\dfrac{7}{3})$.

✓ Verified and validated
Q.3To conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1m each. 100 flower pots have been placed at a distance of 1m from each other along AD, as shown in Fig. 7.12. Niharika runs $\dfrac{1}{4}$ th the distance AD on the 2nd line and posts a green flag. Preet runs $\dfrac{1}{5}$ 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?v
Solution

Taking A as origin, the green flag is on the 2nd line and at $\dfrac14$ of AD, so its coordinates are $(2,25)$. The red flag is on the 8th line and at $\dfrac15$ of AD, so its coordinates are $(8,20)$. Distance $=\sqrt{(8-2)^2+(20-25)^2}=\sqrt{61}$ m. The midpoint is $(\dfrac{2+8}{2},\dfrac{25+20}{2})=(5,22.5)$.

Answer:

The distance between the green and red flags is $\sqrt{61}$ m. Rashmi should post the blue flag at $(5,22.5)$.

✓ Verified and validated
Q.4Find the ratio in which the line segment joining the points (- 3, 10) and (6, - 8) is divided by (- 1, 6).v
Solution

Let the ratio be $m:n$. Using x-coordinate, $-1=\dfrac{6m-3n}{m+n}$. This gives $2n=7m$, so $m:n=2:7$. The y-coordinate gives the same ratio.

Answer:

The ratio is $2:7$.

✓ Verified and validated
Q.5Find 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.v
Solution

A point on the x-axis has y-coordinate 0. Let the ratio be $m:n$. Then $0=\dfrac{5m-5n}{m+n}$, so $m=n$. The point is the midpoint: $(\dfrac{1-4}{2},\dfrac{-5+5}{2})=(-\dfrac32,0)$.

Answer:

The ratio is $1:1$, and the point of division is $(-\dfrac{3}{2},0)$.

✓ Verified and validated
Q.6If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.v
Solution

In a parallelogram, diagonals bisect each other. Midpoint of the diagonal joining $(1,2)$ and $(x,6)$ is $(\dfrac{1+x}{2},4)$. Midpoint of the diagonal joining $(4,y)$ and $(3,5)$ is $(\dfrac72,\dfrac{y+5}{2})$. Equating gives $x=6$ and $y=3$.

Answer:

$x=6$ and $y=3$.

✓ Verified and validated
Q.7Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2, - 3) and B is (1, 4).v
Solution

The centre is the midpoint of diameter AB. If A is $(x,y)$, then $\dfrac{x+1}{2}=2$ and $\dfrac{y+4}{2}=-3$. Therefore $x=3$ and $y=-10$.

Answer:

A is $(3,-10)$.

✓ Verified and validated
Q.8If A and B are (- 2, - 2) and (2, - 4), respectively, find the coordinates of P such that $AP = \dfrac{3}{7} AB$ and P lies on the line segment AB.v
Solution

Since $AP=\dfrac37 AB$, $AP:PB=3:4$. By section formula, $P=(\dfrac{3(2)+4(-2)}{7},\dfrac{3(-4)+4(-2)}{7})=(-\dfrac27,-\dfrac{20}{7})$.

Answer:

$P=(-\dfrac{2}{7},-\dfrac{20}{7})$.

✓ Verified and validated
Q.9Find the coordinates of the points which divide the line segment joining A(- 2, 2) and B(2, 8) into four equal parts.v
Solution

The required points are at $\dfrac14$, $\dfrac12$ and $\dfrac34$ of the way from A to B. Moving from $(-2,2)$ to $(2,8)$ changes coordinates by $(4,6)$, so the points are $(-1,\dfrac72)$, $(0,5)$ and $(1,\dfrac{13}{2})$.

Answer:

$(-1,\dfrac{7}{2})$, $(0,5)$ and $(1,\dfrac{13}{2})$.

✓ Verified and validated
Q.10Find 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 = $\dfrac{1}{2}$ (product of its diagonals)]v
Solution

The diagonals are between $(3,0)$ and $(-1,4)$, and between $(4,5)$ and $(-2,-1)$. Their lengths are $\sqrt{(-4)^2+4^2}=4\sqrt2$ and $\sqrt{(-6)^2+(-6)^2}=6\sqrt2$. Area $=\dfrac12(4\sqrt2)(6\sqrt2)=24$ square units.

Answer:

The area is 24 square units.

✓ Verified and validated
Back: Ch.6
Next: Ch.8