Updated on: 2026-03-31 | Author: Aarti Kulkarni

Geometry (Math 2) Chapter 5 Co Ordinate Geometry Solutions

Q1. Find coordinates of midpoint of segment joining (– 2, 6) and (8, 2)

Let A(x1, y1) = A(– 2, 6), B(x2, y2) = B(8, 2), C(x, y) be the midpoint of seg AB.
∴ x1 = – 2, y1 = 6, x2 = 8, y2 = 2
By midpoint formula,

C(x, y) = `((x_1 + x_2)/2, (y_1 + y_2)/2)`

= `((-2 + 8)/2, (6 + 2)/2)`

= `(6/2, 8/2)`

= (3, 4)

∴ Coordinates of midpoint of segment joining (– 2, 6) and (8, 2) are (3, 4).

Q2. Find coordinates of midpoint of the segment joining points (0, 2) and (12, 14)

Let A(x1, y1) = A(0, 2), B (x2, y2) = B(12, 14)

Let the co-ordinates of the midpoint be P(x, y).

∴ By midpoint formula,
x = `(x_1 + x_2)/2 = (0 + 12)/2` = 6
y = `(y_1 + y_2)/2 = (2 + 14)/2 = 16/2` = 8
∴ The co-ordinates of the midpoint of the segment joining (0, 2) and (12, 14) are (6, 8).

Q3. Find coordinates of the midpoint of a segment joining point A(–1, 1) and point B(5, –7) Solution: Suppose A(x 1 , y 1 ) and B(x 2 , y 2 ) x 1 = –1, y 1 = 1 and x 2 = 5, y 2 = –7 Using midpoint formula, ∴ Coordinates of midpoint of segment AB = `((x_1 + x_2)/2, (y_1+ y_2)/2)` = `(square/2, square/2)` ∴ Coordinates of the midpoint = `(4/2, square/2)` ∴ Coordinates of the midpoint = `(2, square)`

Suppose A(x1, y1) and B(x2, y2)

x1 = –1, y1 = 1 and x2 = 5, y2 = –7
Using midpoint formula,
∴ Coordinates of midpoint of segment AB

= `((x_1 + x_2)/2, (y_1+ y_2)/2)`

= `((-1 + 5)/2, (1 - 7)/2)`

∴ Coordinates of the midpoint = `(4/2, (-6)/2)`
∴ Coordinates of the midpoint = (2, – 3)

Q4. Find distance between point A(–1, 1) and point B(5, –7): Solution: Suppose A(x 1 , y 1 ) and B(x 2 , y 2 ) x 1 = –1, y 1 = 1 and x 2 = 5, y 2 = – 7 Using distance formula, d(A, B) = `sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2` ∴ d(A, B) = `sqrt(square +[(-7) + square]^2` ∴ d(A, B) = `sqrt(square)` ∴ d(A, B) = `square`

Suppose A(x1, y1) and B(x2, y2)

x1 = –1, y1 = 1 and x2 = 5, y2 = – 7
Using distance formula,

d(A, B) = `sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2`

∴ d(A, B) = `sqrt([5 - (-1)]^2 + [(-7) + -1]^2`
∴ d(A, B) = `sqrt(6^2 + (-8)^2`
∴ d(A, B) = `sqrt(36 + 64)`
∴ d(A, B) = `sqrt(100)`
∴ d(A, B) = 10 units

Q5. Find distance between point A(– 3, 4) and origin O

5 cm

Let A(x1, y1) = A( -3, 4) and O(x2, y2) = O(0, 0)
Here, x1 = -3, y1 = 4, x2 = 0, y2 = 0
By distance formula,

d(A, O) = `sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2)`

∴ d(A, O) = `sqrt([0 - (-3)]^2 + (0- 4)^2)`
∴ d(A, O) = `sqrt(9 + 16)`
∴ d(A, O) = `sqrt(25)`
∴ d(A, O) = 5 cm

Q6. Find distance between point A(7, 5) and B(2, 5)

Let A(x1, y1) = A(7, 5) and B(x2, y2) = B(2, 5)
∴ By distance formula,

d(A, B) = `sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2`

= `sqrt((2- 7)^2 + (5 - 5)^2`

= `sqrt((-5)^2 + 0^2)`

= `sqrt(25)`

= 5 cm

∴ The distance between points A and B is 5 cm.

Q7. Find distance between point Q(3, – 7) and point R(3, 3) Solution: Suppose Q(x 1 , y 1 ) and point R(x 2 , y 2 ) x 1 = 3, y 1 = – 7 and x 2 = 3, y 2 = 3 Using distance formula, d(Q, R) = `sqrt(square)` ∴ d(Q, R) = `sqrt(square - 100)` ∴ d(Q, R) = `sqrt(square)` ∴ d(Q, R) = `square`

Suppose Q(x1, y1) and point R(x2, y2)

x1 = 3, y1 = – 7 and x2 = 3, y2 = 3
Using distance formula,

d(Q, R) = `sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2`

= `sqrt((3 - 3)^2 - [3 - (- 7)]^2`

= `sqrt(0^2 + (10)^2)`

∴ d(Q, R) = `sqrt(0 - 100)`
∴ d(Q, R) = `sqrt(100)`
∴ d(Q, R) = 10

Q8. Find distance between points O(0, 0) and B(– 5, 12)

Let O(x1, y1) = O(0, 0) and B(x2, y2) = B(– 5, 12)
∴ x1 = 0, y1 = 0, x2 = – 5, y2 = 12
By distance formula,

d(O, B) = `sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2`

= `sqrt((-5 - 0)^2 + (12 - 0)^2`

= `sqrt((-5)^2 + 12^2`

= `sqrt(25 + 144)`

= `sqrt(169)`

∴ d(O, B) = 13 units
∴ The distance between the points O and B is 13 units.

Q9. Find distance CD where C(– 3a, a), D(a, – 2a)

Let C(x1, y1) and D(x2, y2) be the given points

∴ x1 = – 3a, y1 = a, x2 = a, y2 = – 2a
By distance formula,

d(C, D) = `sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2`

= `sqrt(["a" - (-3"a")]^2 + (-2"a" - "a")^2`

= `sqrt(("a" + 3"a")^2 + (-2"a" - "a")^2`

= `sqrt((4"a")^2 + (-3"a")^2`

= `sqrt(16"a"^2 + 9"a"^2)`

= `sqrt(25"a"^2)`

∴ d(C, D) = 5a units

Q10. Find distance of point A(6, 8) from origin

Let A(x1, y1) = A(6, 8), O(x2, y2) = O(0, 0)
∴ x1 = 6, y1 = 8, x2 = 0, y2 = 0
By distance formula,

d(A, O) =`sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2`

= `sqrt((0 - 6)^2 + (0 - 8)^2`

= `sqrt(36 + 64)`

= `sqrt(100)`

= 10 cm

∴ The distance of point A(6, 8) from origin is 10 cm.

Q11. Find the coordinates of centroid of a triangle whose vertices are (4, 7), (8, 4) and (7, 11)

Let A(x1, y1) = A(4, 7), B(x2, y2) = B(8, 4), C(x3, y3) = C(7, 11)
∴ By centroid formula,

x = `(x_1 + x_2 + x_3)/3`

= `(4 + 8 + 7)/3`

= `19/3`

y = `(y_1 + y_2 + y_3)/3`

= `(7 + 4 + 11)/3`

= `22/3`

∴ The co-ordinates of the centroid are `(19/3, 22/3)`

Q12. Find the coordinates of midpoint of segment joining (22, 20) and (0, 16)

Let A(x1, y1) = A(22, 20), B(x2, y2) = B(0, 16)

Let the co-ordinates of the midpoint be P(x, y).

∴ By midpoint formula,

x = `(x_1 + x_2)/2`

= `(22 + 0)/2`

= 11

y = `(y_1 + y_2)/2`

= `(20 + 16)/2`

= `36/2`

= 18

∴ The co-ordinates of the midpoint of the segment joining (22, 20) and (0, 16) are (11, 18).

Q13. Find the coordinates of the point of intersection of the graph of the equation x = 2 and y = – 3

The coordinates of the point of intersection of the graph of the equation x = 2 and y = – 3 are (2, – 3).

Q14. Find the ratio in which Y-axis divides the point A(3, 5) and point B(– 6, 7). Find the coordinates of the point

Let C be a point on Y-axis which divides seg AB in the ratio m : n

Point C lies on the Y-axis.

∴ its X-coordinate is 0.
Let C = (0, y)

Here,

A(x1, y1) = A(3, 5)
B(x2, y2) = B(– 6, 7)
By Section formula,

x = `("m"x_2 + "n"x_1)/("m" + "n")`

∴ 0 = `(-6"m" + 3"n")/("m" + "n")`
∴ – 6m + 3n = 0
∴ 3n = 6m
∴ `"m"/"n" = 3/6`
∴ `"m"/"n" = 1/2` ......(i)
∴ m : n = 1 : 2
By section formula,

y = `("m"y_2 + "n"y_1)/("m" + "n")`

y = `(7"m" + 5"n")/("m" + "n")`

= `(7"m" + 5(2"m"))/("m" + 2"m")` ......[From (i), n = 2m]

= `(7"m" + 10"m")/(3"m")`

= `(17"m")/(3"m")`

∴ Y-axis divides the seg AB in the ratio 1 : 2 and the co-ordinates of that point is `(0, 17/3)`.

Q15. From the figure given alongside, find the length of the median AD of triangle ABC. Complete the activity. Solution: Here A(–1, 1), B(5, – 3), C(3, 5) and suppose D(x, y) are coordinates of point D. Using midpoint formula, x = `(5 + 3)/2` ∴ x = `square` y = `(-3 + 5)/2` ∴ y = `square` Using distance formula, ∴ AD = `sqrt((4 - square)^2 + (1 - 1)^2` ∴ AD = `sqrt((square)^2 + (0)^2` ∴ AD = `sqrt(square)` ∴ The length of median AD = `square`

Here A(–1, 1), B(5, – 3), C(3, 5) and suppose D(x, y) are coordinates of point D. D is the midpoint of seg BC.

Using midpoint formula,

x = `(x_1 + x_2)/2`

x = `(5 + 3)/2`

∴ x = `8/2`
∴ x = 4

y = `(y_1 + y_2)/2`

y = `(-3 + 5)/2`

∴ y = `2/2`
∴ y = 1
Using distance formula,
∴ AD = `sqrt((4 - (-1))^2 + (1 - 1)^2`
∴ AD = `sqrt((5)^2 + (0)^2`
∴ AD = `sqrt(25)`
∴ The length of median AD = 5 cm

Q16. If A(5, 4), B(–3, –2) and C(1, –8) are the vertices of a ∆ABC. Segment AD is median. Find the length of seg AD:

Diagram: Refer textbook

Q17. If point P(1, 1) divide segment joining point A and point B(–1, –1) in the ratio 5 : 2, then the coordinates of A are ______

(6, 6)

Let A(x1, y1) and B(x2, y2) = B(-1, -1)
P(x, y) = P(1, 1) divides AB in ratio 5 : 2.
∴ x1 = 1, y1 = 1, x2 = –1, y2 = –1, a = 5, b = 2.
∴ By section formula,
∴ x = `(ax_2 + bx_1)/(a + b)`
∴ `1 = (5(-1) + 2x_1)/(5 + 2)`
∴ 7 = -5 + 2x1
∴ 2x1 = 7 + 5
∴ 2x1 = 12
∴ x1 = `12/2`
∴ x1 = 6
∴ Co-ordinates of A are (6, 6).

Q18. If point P divides segment AB in the ratio 1 : 3 where A(– 5, 3) and B(3, – 5), then the coordinates of P are ______

(– 3, 1)

Let A(x1, y1) = A(-5, 3) and B(x2, y2) = B(3, -5),
a : b = 1 : 3
∴ x1 = -5, y1 = 3, x2 = 3, y2 = –5, a = 1, b = 3.
∴ By section formula,

∴ x = `(ax_2 + bx_1)/(a + b)`

∴ x = `(1(3) + 3(-5))/(1 + 3)`

∴ x = `(3 - 15)/(4)`

∴ x = `(- 12)/(4)`

∴ x = -3

∴ y = `(ay_2 + by_1)/(a + b)`

∴ y = `(1(-5) + 3(3))/(1 + 3)`

∴ y = `(-5 + 9)/(4)`

∴ y = `(4)/(4)`

∴ y = 1
∴ Co-ordinates of P are (-3, 1).

Q19. If point P is midpoint of segment joining point A(– 4, 2) and point B(6, 2), then the coordinates of P are ______

(1, 2)

A(x1, y1) = A( –4, 2), B(x2, y2) = B(6, 2)
Here, x1 = – 4, y1 = 2, x2 = 6, y2 = 2
∴ Co-ordinates of the midpoint of seg AB

= `((x_1 + x_2)/2, (y_1 + y_2)/2)`

= `((-4 + 6)/2, (2 + 2)/2)`

= `(2/2, 4/2)`

= (1, 2)

Q20. If segment AB is parallel Y-axis and coordinates of A are (1, 3), then the coordinates of B are ______

(1, – 3)

Since, seg AB || Y-axis.

∴ x co-ordinate of all points on seg AB will be the same.
x co-ordinate of A (1, 3) = 1
x co-ordinate of B (1, – 3) = 1
∴ (1, – 3) is option correct.

Q21. If the distance between point L(x, 7) and point M(1, 15) is 10, then find the value of x

Let L(x1, y1) = L(x, 7) and M (x2, y2) = M(1, 15)
x1 = x, y1 = 7, x2 = 1, y2 = 15
By distance formula,

d(L, M) = `sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2`

∴ d(L, M) = `sqrt((1 - x)^2 + (15 - 7)^2`
∴ 10 = `sqrt((1 - x)^2 + 8^2`
∴ 100 = (1 – x)2 + 64 ......[Squaring both sides]
∴ (1 – x)2 = 100 – 64
∴ (1 – x)2 = 36
∴ 1 – x = `+- sqrt(36)` .....[Taking square root of both sides]
∴ 1 – x = ± 6
∴ 1 – x = 6 or 1 – x = – 6
∴ x = – 5 or x = 7
∴ The value of x is – 5 or 7.

Q22. If the length of the segment joining point L(x, 7) and point M(1, 15) is 10 cm, then the value of x is ______

7 or – 5

Here, x1 = x, y1 = 7, x2 = 1, y2 = 15
By distance formula,

d(L, M) = `sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2)`

∴ d(L, M) = `sqrt((1 - x)^2 + (15 - 7)^2)`
∴ 10 = `sqrt((1 - x)^2 + 8^2)`
∴ 100 = (1 - x)2 + 64 ...[Squaring both sides]
∴ (1 - x)2 = 100 - 64
∴ (1 - x)2 = 36
∴ 1 - x = `+-sqrt(36)` ...[Taking square root of both sides]
∴ 1 - x = `+-6`
∴ 1 - x = 6 or 1 - x = -6
∴ x = -5 or x = 7
∴ The value of x is -5 or 7.

Q23. If the point P (6, 7) divides the segment joining A(8, 9) and B(1, 2) in some ratio, find that ratio Solution: Point P divides segment AB in the ratio m: n. A(8, 9) = (x 1 , y 1 ), B(1, 2 ) = (x 2 , y 2 ) and P(6, 7) = (x, y) Using Section formula of internal division, ∴ 7 = `("m"(square) - "n"(9))/("m" + "n")` ∴ 7m + 7n = `square` + 9n ∴ 7m – `square` = 9n – `square` ∴ `square` = 2n ∴ `"m"/"n" = square`

Point P divides segment AB in the ratio m : n.

A(8, 9) = (x1, y1), B(1, 2 ) = (x2, y2) and P(6, 7) = (x, y)
Using Section formula of internal division,

y = `("m"y_2 + "n"y_1)/("m" + "n")`

∴ 7 = `("m"(2) - "n"(9))/("m" + "n")`
∴ 7m + 7n = 2m + 9n
∴ 7m – 2m = 9n – 7n
∴ 5m = 2n
∴ `"m"/"n"` = `bb(2/5)`

Q24. If the sum of X-coordinates of the vertices of a triangle is 12 and the sum of Y-coordinates is 9, then the coordinates of centroid are ______

(4, 3)

Let P(x1, y1), Q(x2, y2), R(x3, y3) be the co-ordinates of the vertices of a triangle.

Given that x1 + x2 + x3 = 12 and y1 + y2 + y3 = 9
∴ Co-ordinates of the centroid S(x, y) are
`x = (x_1 + x_2 + x_3)/3 "and" y = (y_1 + y_2 + y_3)/3`
`x = (12)/3 "and" y = (9)/3`
`x = 4 "and" y = 3`
∴ S(x, y) = (4, 3).

Q25. Point P(– 4, 6) divides point A(– 6, 10) and B(m, n) in the ratio 2:1, then find the coordinates of point B

By section formula

– 4 = `(2 xx "m" + 1 xx (-6))/(2 + 1)`

∴ – 4 = `(2"m" - 6)/3`
∴ –12 = 2m – 6
∴ 2m = – 6
∴ m = – 3

6 = `(2 xx "n" + 1 xx 10)/(2 + 1)`

∴ 6 = `(2"n" + 10)/3`
∴ 18 = 2n + 10
∴ 2n = 8
∴ n = 4

Co-ordinates of point B are (– 3, 4).

Q26. Point P is midpoint of segment AB where A(– 4, 2) and B(6, 2), then the coordinates of P are ______

(1, 2)

A(x1, y1) = A( –4, 2), B(x2, y2) = B(6, 2)
Here, x1 = – 4, y1 = 2, x2 = 6, y2 = 2
∴ Co-ordinates of the midpoint of seg AB

= `((x_1 + x_2)/2, (y_1 + y_2)/2)`

= `((-4 + 6)/2, (2 + 2)/2)`

= `(2/2, 4/2)`

= (1, 2)

Q27. Seg OA is the radius of a circle with centre O. The coordinates of point A is (0, 2) then decide whether the point B(1, 2) is on the circle?

Diagram: Refer textbook

Q28. Show that A(1, 2), (1, 6), C(1 + 2 `sqrt(3)`, 4) are vertices of a equilateral triangle

Distance between two points = `sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2`

By distance formula,

AB = `sqrt((1- 1)^2 + (6 - 2)^2`

= `sqrt(0^2 + 4^2)`

= `sqrt(4^2)`

= 4 ......(i)

BC = `sqrt((1 + 2sqrt(3) - 1)^2 + (4 - 6)^2`

= `sqrt((2sqrt(3))^2 + (-2)^2`

= `sqrt(12 + 4)`

= `sqrt(16)`

= 4 .....(ii)

AC = `sqrt((1 + 2sqrt(3) - 1)^2 + (4 -2)^2`

= `sqrt((2sqrt(3))^2 + 2^2`

= `sqrt(12 + 4)`

= `sqrt(16)`

= 4 ......(iii)

∴ AB = BC = AC ......[From (i), (ii) and (iii)]
∴ ∆ABC is an equilateral triangle.
∴ Points A, B and C are the vertices of an equilateral triangle.

Q29. Show that P(– 2, 2), Q(2, 2) and R(2, 7) are vertices of a right angled triangle

Distance between two points = `sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2`

By distance formula,

PQ = `sqrt([2 - (-2)]^2 + (2 - 2)^2`

= `sqrt((2 + 2)^2 + (0)^2`

= `sqrt((4)^2`

= 4 .....(i)

QR = `sqrt((2 - 2)^2 + (7 - 2)^2`

= `sqrt((0)^2 + (5)^2`

= `sqrt((5)^2`

= 5 ......(ii)

PR = `sqrt([2 -(-2)]^2 + (7 - 2)^2`

= `sqrt((2 + 2)^2 + (5)^2`

= `sqrt((4)^2 + (5)^2`

= `sqrt(16 + 25)`

= `sqrt(41)`

Now, PR2 = `(sqrt(41))^2` = 41 ......(iii)

Consider, PQ2 + QR2

= 42 + 52

= 16 + 25

= 41 ......[From (i) and (ii)]

∴ PR2 = PQ2 + QR2 ......[From (iii)]
∴ ∆PQR is a right angled triangle. ......[Converse of Pythagoras theorem]
∴ Points P, Q, and R are the vertices of a right angled triangle.

Q30. Show that points A(– 4, –7), B(–1, 2), C(8, 5) and D(5, – 4) are the vertices of a parallelogram ABCD

We know that, slope of line = `(y_2 - y_1)/(x_2 - x_1)`

Slope of side AB = `(2 - (-7))/(-1 - (-4))`

= `(2 + 7)/(-1 + 4)`

= `9/3`

= 3 ......(i)

Slope of side BC = `(5 - 2)/(8 - (-1))`

= `3/(8 + 1)`

= `3/9`

= `1/3` ......(ii)

Slope of side CD = `(-4 - 5)/(5 - 8)`

= `(-9)/(-3)`

= 3 .....(iii)

Slope of side AD = `(-4 - (-7))/(5 - (-4))`

= `(-4 + 7)/(5 + 4)`

=`3/9`

= `1/3` ......(iv)

∴ Slope of side AB = Slope of side CD ......[From (i) and (iii)]
∴ Side AB || side CD
∴ Slope of side BC = Slope of side AD .......[From (ii) and (iv)]
∴ Side BC || side AD

Both the pairs of opposite sides of ABCD are parallel

∴ ▢ABCD is a parallelogram.
∴ Points A(– 4, – 7), B(– 1, 2), C(8, 5) and D(5, – 4) are the vertices of a parallelogram.

Q31. Show that the point (0, 9) is equidistant from the points (– 4, 1) and (4, 1)

Let P(x1, y1) = P(0, 9), Q(x2, y2) = Q(– 4, 1), R(x3, y3) = R(4, 1)
By distance formula,

d(P, Q) = `sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2`

= `sqrt([(-4) - 0]^2 + (1 - 9)^2`

= `sqrt((-4)^2 + (-8)^2`

= `sqrt(16 + 64)`

= `sqrt(80)`

= `4sqrt(5)`

And

d(P, R) = `sqrt((x_3 - x_1)^2 + (y_3 - y_1)^2`

= `sqrt((4 - 0)^2 + (1 - 9)^2`

= `sqrt(4^2 + (-8)^2`

= `sqrt(16 + 64)`

= `sqrt(80)`

= `4sqrt(5)`

Here, d(P, Q) = d(P, R)
∴ The point (0, 9) is equidistant from (– 4, 1) and (4, 1).

Q32. Show that the point (11, – 2) is equidistant from (4, – 3) and (6, 3)

Let P(x1, y1) = P(11, – 2), Q(x2, y2) = Q(4, – 3), R(x3, y3) = R(6, 3)
By distance formula,

d(P, Q) = `sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2`

= `sqrt((4 - 11)^2 + [-3 - (-2)]^2`

= `sqrt((-7)^2 + (-1)^2`

= `sqrt(49 + 1)`

= `sqrt(50)`

= `5sqrt(2)`

And

d(P, R) = `sqrt((x_3 - x_1)^2 + (y_3 - y_1)^2`

= `sqrt((6 - 11)^2 + [3 - (-2)]^2`

= `sqrt((-5)^2 + (5)^2`

= `sqrt(25 + 25)`

= `sqrt(50)`

= `5sqrt(2)`

Here, d(P, Q) = d(P, R)
∴ Point (11, – 2) is equidistant from (4, – 3) and (6, 3).

Q33. Show that the points (0, –1), (8, 3), (6, 7) and (– 2, 3) are vertices of a rectangle.

Let the points be P(0, –1), Q(8, 3), R(6, 7), S(–2, 3)

Distance between two points= `sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2`

∴ By distance formula,

d(P, Q) = `sqrt((8 - 0)^2 + [3 - (-1)]^2`

= `sqrt((8 - 0)^2 + (3 + 1)^2`

= `sqrt(8^2 + 4^2)`

= `sqrt(64 + 16)`

= `sqrt(80)` ......(i)

d(Q, R) = `sqrt((6 - 8)^2 + (7 - 3)^2`

= `sqrt((-2)^2 + (4)^2`

= `sqrt(4 + 16)`

= `sqrt(20)` ......(ii)

d(R, S) = `sqrt([(-2) - 6]^2 + (3 - 7)^2`

= `sqrt((-8)^2 + (-4)^2`

= `sqrt(64 + 16)`

=`sqrt(80)` ......(iii)

d(P, S) = `sqrt([(-2) - 0]^2 + [3 - (-1)^2]`

= `sqrt((-2)^2+ (3+ 1)^2`

= `sqrt((-2)^2 + 4^2`

= `sqrt(4 + 16)`

= `sqrt(20)` ......(iv)

In ▢PQRS,

∴ side PQ = side RS .......[From (i) and (iii)]
side QR = side PS ......[From (ii) and (iv)]
∴ ▢PQRS is a parallelogram ......[A quadrilateral is a parallelogram, if both the pairs of its opposite sides are congruent]

d(P, R) = `sqrt((6 - 0)^2 + [7 - (-1)]^2`

= `sqrt((6 - 0)^2 + (7 + 1)^2`

= `sqrt(6^2 + 8^2)`

= `sqrt(36 + 64)`

= `sqrt(100)`

= 10 ......(iv)

d(Q, S) = `sqrt([(-2) - 8]^2 + [3 - 3]^2`

= `sqrt((-10)^2 + (0)^2`

= `sqrt(100 + 0)`

= `sqrt(100)`

= 10 ......(vi)

In parallelogram PQRS,

PR = QS .......[From (v) and (vi)]
∴ ▢PQRS is a rectangle. .......[A parallelogram is a rectangle if its diagonals are equal]

Q34. Show that the points (2, 0), (– 2, 0) and (0, 2) are vertices of a triangle. State the type of triangle with reason

Answer

Let the points be P(2, 0), Q(– 2, 0) and R(0, 2)

Distance between two points = `sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2`

By distance formula,

d(P, Q) = `sqrt([(-2) - 2]^2 + (0 - 0)^2`

= `sqrt((-4)^2 + (0)^2`

= `sqrt(16 + 0)`

= 4 .....(i)

d(Q, R) = `sqrt([0 - (-2)]^2 + (2 - 0)^2`

= `sqrt((2)^2 + (2)^2`

= `sqrt(4 + 4)`

= `sqrt(8)` ......(ii)

d (P, R) = `sqrt((0 -2)^2 + (2 - 0)^2`

= `sqrt((- 2)^2 + (2)^2`

= `sqrt(4 + 4)`

= `sqrt(8)` ......(iii)

On adding (ii) and (iii),

d(P, Q) + d(Q, R) = `4 + sqrt(8)`

`4 + sqrt(8) > sqrt(8)`

∴ d(P, Q) + d(Q, R) > d(P, R)
∴ Points P, Q, R are non colinear points.

We can construct a triangle through 3 non collinear points.

∴ The segment joining the given points form a triangle.
Since P(Q, R) = P(P, R)
∴ ∆PQR is an isosceles triangle.
∴ The segment joining the points (2, 0), (– 2, 0) and (0, 2) will form an isosceles triangle.

Q35. The coordinates of diameter AB of a circle are A(2, 7) and B(4, 5), then find the coordinates of the centre

Let C(x, y) be the centre of the circle,

A(x1, y1) = A(2, 7), B(x2, y2) = B(4, 5)
∴ x1 = 2, y1 = 7, x2 = 4, y2 = 5

C is the mid-point of seg AB.

∴ By midpoint formula,

C(x, y) = `((x_1 + x_2)/2, (y_1 + y_2)/2)`

= `((2 + 4)/2, (7 + 5)/2)`

= `(6/2, 12/2)`

∴ C(x, y) = C(3, 6)
∴ The co-ordinates of the centre of the circle are (3, 6).

Q36. The coordinates of the vertices of a triangle ABC are A (–7, 6), B(2, –2) and C(8, 5). Find coordinates of its centroid. Solution: Suppose A(x 1 , y 1 ) and B(x 2 , y 2 ) and C(x 3 , y 3 ) x 1 = –7, y 1 = 6 and x 2 = 2, y 2 = –2 and x 3 = 8, y 3 = 5 Using Centroid formula ∴ Coordinates of the centroid of a traingle ABC = `((x_1 + x_2 + x_3)/3, (y_1 + y_2 + y_3)/3)` = `(square/3, square/3)` ∴ Coordinates of the centroid of a triangle ABC = `(3/3, square)` ∴ Coordinates of the centroid of a triangle ABC = `(1 , square)`

Suppose A(x1, y1) and B(x2, y2) and C(x3, y3)

x1 = –7, y1 = 6 and x2 = 2, y2 = –2 and x3 = 8, y3 = 5
Using Centroid formula
∴ Coordinates of the centroid of a traingle

ABC = `((x_1 + x_2 + x_3)/3, (y_1 + y_2 + y_3)/3)`

= `((-7 + 2 + 8)/3, (6 - 2 + 5)/3)`

∴ Coordinates of the centroid of a triangle ABC = `(3/3, 9/3)`
∴ Coordinates of the centroid of a triangle ABC = (1 , 3)

Q37. The distance between point P(2, 2) and Q(5, x) is 5 cm, then the value of x ______

6

Let P(x1, y1) = P( 2, 2) and Q(x2, y2) = Q(5, x)
Here, x1 = 2, y1 = 2, x2 = 5, y2 = x
By distance formula,

d(P, Q) = `sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2)`

∴ 5 = `sqrt((5 - 2)^2 + (x - 2)^2)`
∴ 5 = `sqrt(9 + x^2 - 4x + 4)`
∴ 52 = x2 - 4x + 13 ...[Squaring both sides]
∴ 25 = x2 - 4x + 13
∴ x2 - 4x + 13 - 25 = 0
∴ x2 - 4x - 12 = 0
∴ (x - 6) (x + 2) = 0
∴ x - 6 = 0 or x + 2 = 0
∴ x = 6 or x = -2

Q38. The distance between points P(–1, 1) and Q(5, –7) is ______

10 cm

Let P(x1, y1) = P( -1, 1) and Q(x2, y2) = Q(5, -7)
Here, x1 = -1, y1 = 1, x2 = 5, y2 = -7
By distance formula,

d(P, Q) = `sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2)`

∴ d(P, Q) = `sqrt([5 - (-1)]^2 + (-7- 1)^2)`
∴ d(P, Q) = `sqrt(36 + 64)`
∴ d(P, Q) = `sqrt(100)`
∴ d(P, Q) = 10 cm

Q39. The point Q divides segment joining A(3, 5) and B(7, 9) in the ratio 2 : 3. Find the X-coordinate of Q

Let the co-ordinates of point Q be (x, y) and A (x1, y1), B (x2, y2) be the given points.

Here, x1 = 3, y1 = 5, x2 = 7, y2 = 9, m = 2, n = 3
∴ By section formula,

x = `("m"x_2 + "n"x_1)/("m" + "n")`

= `(2(7) + 3(3))/(2 + 3)`

= `(14 + 9)/5`

= `23/5`

y = `("m"y_2 + "n"y_1)/("m" + "n")`

= `(2(9) + 3(5))/(2 + 3)`

= `(18 + 15)/5`

= `33/5`

∴ The co-ordinates of point Q are `(23/5, 33/5)`.

Q40. The points (7, – 6), (2, k) and (h, 18) are the vertices of triangle. If (1, 5) are the coordinates of centroid, find the value of h and k

Let (7, – 6) ≡ (x1, y1), (2, k) ≡ (x2, y2), (h, 18) ≡ (x3, y3) be the three vertices of the triangle.

(1, 5) are coordinates of centroid.

∴ By Centroid formula,

1 = `(x_1 + x_2 + x_3)/3`

∴ 1 = `(7 + 2 + "h")/3`
∴ 3 = 9 + h
∴ h = – 6

5 = `(y_1 + y_2 + y_3)/3`

∴ 5 = `(-6 + "k" + 18)/3`
∴ 15 = 12 + k
∴ k = 3

Q41. Using distance formula decide whether the points (4, 3), (5, 1), and (1, 9) are collinear or not.

Let A(x1, y1) = A(4, 3), B(x2, y2) = B(5, 1), C(x3, y3) = C(1, 9)
∴ d(A, B) = `sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2`

= `sqrt((5 -4)^2 + (1 - 3)^2`

= `sqrt(1^2 + (-2)^2`

= `sqrt(1+ 4)`

= `sqrt(5)` ......(i)

∴ d(B, C) = `sqrt((x_3 - x_2)^2 + (y_3 - y_2)^2`

= `sqrt((1 - 5)^2 + (9 - 1)^2`

= `sqrt((-4)^2 + 8^2`

= `sqrt(16 + 64)`

=`sqrt(80)`

= `4sqrt(5)` ......(ii)

∴ d(A, C) = `sqrt((x_3 -x_1)^2 + (y_3 - y_2)^2`

= `sqrt((1 - 4)^2 + (9 - 3)^2`

= `sqrt((-3)^2 + 6^2`

= `sqrt(9 + 36)`

= `sqrt(45)`

= `3sqrt(5)` .....(iii)

`sqrt(5) + 3sqrt(5) = 4sqrt(5)`
∴ d(A, B) + d(A, C) = d(B, C) ......[From (i), (ii) and (iii)]
∴ Points A, B, C are collinear.

Q42. What are the coordinates of origin?

Co-ordinates of origin are (0, 0).

Q43. Write the X-coordinate and Y-coordinate of point P(– 5, 4)

Answer

X-co-ordinate of point P(– 5, 4) is –5,

Y-co-ordinate of point P(– 5, 4) is 4

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