# Frank Solutions for Chapter 12 Distance and Section Formula Class 10 ICSE Mathematics

### Exercise 12.1

1. Find the distance between the following pairs of points in the coordinate plane:

(a) (5, -2) and (1, 5)

(b) (1, 3) and (3, 9)

(c) (7, -7) and (2, 5)

(d) (4, 1) and (-4, 5)

(e) (13, 7) and (4, -5)

(a) (5, -2) and (1, 5)

A = (5, – 2)

B = (1, 5)

AB = √((x2 – x1)2 + (y2 – y1)2

Where, x1 = x coordinate of A = 5

y1 = y coordinate of A = – 2

x2 = x coordinate of B = 1

y2 = y coordinates of B = 5

= √((1 – 5)2 + (5 + 2)2)

= √((4)2 + (7)2)

= √(16 + 49)

= √65 units

(b) (1, 3) and (3, 9)

A = (1, 3), B = (3, 9)

AB = √((x2 – x1)2 + (y2 – y1)2

Where, x1 = x coordinate of A = 1

y1 = y coordinate of A = 3

x2 = x coordinate of B = 3

y2 = y coordinates of B = 9

= √((3 – 1)2 + (9 – 3)2)

= √((2)2 + (6)2)

= √(4 + 36)

= √40 units

= 2√10 units

(c) (7, -7) and (2, 5)

A = (7, -7), B = (2, 5)

AB = √((x2 – x1)2 + (y2 – y1)2

Where, x1 = x coordinate of A = 7

y1 = y coordinate of A = -7

x2 = x coordinate of B = 2

y2 = y coordinates of B = 5

= √((2 – 7)2 + (5 + 7)2)

= √((-5)2 + (12)2)

= √(25 + 144)

= √169 units

= 13 units

(d) (4, 1) and (-4, 5)

A = (4, 1), B = (-4, 5)

AB = √((x2 – x1)2 + (y2 – y1)2

Where, x1 = x coordinate of A = 4

y1 = y coordinate of A = 1

x2 = x coordinate of B = -4

y2 = y coordinates of B = 5

= √((4 + 4)2 + (1 – 5)2)

= √((8)2 + (4)2)

= √(64 + 16)

= √80 units

= 4√5 units

(e) (13, 7) and (4, -5)

A = (13, 7), B = (4, -5)

AB = √((x2 – x1)2 + (y2 – y1)2

Where, x1 = x coordinate of A = 13

y1 = y coordinate of A = 7

x2 = x coordinate of B = 4

y2 = y coordinates of B = -5

= √((4 – 13)2 + (- 5 – 7)2)

= √((-9)2 + (-12)2)

= √(81 + 144)

= √225 units

= 15 units

2. Find the distances of the following points from the origin.

(a) (5, 12)

(b) (6, 8)

(c) (8, 15)

(d) (0, 11)

(e) (13, 0)

(a) (5, 12)

Origin A = (0, 0), B = (5, 12)

AB = √((x2 – x1)2 + (y2 – y1)2

Where, x1 = x coordinate of A = 0

y1 = y coordinate of A = 0

x2 = x coordinate of B = 5

y2 = y coordinates of B = 12

= √((5 – 0)2 + (12 – 0)2)

= √((5)2 + (12)2)

= √(25 + 144)

= √169 units

= 13 units

(b) (6, 8)

Origin A = (0, 0), B = (6, 8)

AB = √((x2 – x1)2 + (y2 – y1)2

Where, x1 = x coordinate of A = 0

y1 = y coordinate of A = 0

x2 = x coordinate of B = 6

y2 = y coordinates of B = 8

= √((6 – 0)2 + (8 – 0)2)

= √((6)2 + (8)2)

= √(36 + 64)

= √100 units

= 10 units

(c) (8, 15)

Origin A = (0, 0), B = (8, 15)

AB = √((x2 – x1)2 + (y2 – y1)2

Where, x1 = x coordinate of A = 0

y1 = y coordinate of A = 0

x2 = x coordinate of B = 8

y2 = y coordinates of B = 15

= √((8 – 0)2 + (15 – 0)2)

= √((8)2 + (15)2)

= √(64 + 225)

= √289 units

= 17 units

(d) (0, 11)

Origin A = (0, 0), B = (0, 11)

AB = √((x2 – x1)2 + (y2 – y1)2

Where, x1 = x coordinate of A = 0

y1 = y coordinate of A = 0

x2 = x coordinate of B = 0

y2 = y coordinates of B = 11

= √((0 – 0)2 + (11 – 0)2)

= √((0)2 + (11)2)

= √(0 + 121)

= √121 units

= 11 units

(e) (13, 0)

Origin A = (0, 0), B = (13, 0)

AB = √((x2 – x1)2 + (y2 – y1)2

Where, x1 = x coordinate of A = 0

y1 = y coordinate of A = 0

x2 = x coordinate of B = 13

y2 = y coordinates of B = 0

= √((13 – 0)2 + (0 – 0)2)

= √((13)2 + (0)2)

= √(169 + 0)

= √169 units

= 13 units

3. Find the distance between the following points.

(a) (p + q, p – q) and (p – q, p – q)

(b) (sin Î¸, cos Î¸) and (cos Î¸, -sin Î¸)

(c) (sec Î¸, tan Î¸) and (-tan Î¸, sec Î¸)

(d) (sin Î¸ – cosec Î¸, cos Î¸ – cot Î¸) and (cos Î¸ – cosec Î¸, – sin Î¸ – cot Î¸)

(a) (p + q, p – q) and (p – q, p – q)

A (p + q, p – q), B (p – q, p – q)

AB = √((x2 – x1)2 + (y2 – y1)2

Where, x1 = x coordinate of A = p + q

y1 = y coordinate of A = p – q

x2 = x coordinate of B = p – q

y2 = y coordinates of B = p – q

= √[((p – q) – (p + q))2 + ((p – q) – (p – q))2]

= √[(p – q – p – q)2 + (p – q – p + q)2]

= √[(-2q)2 + (0)2]

= √4q units

(b) (sin Î¸, cos Î¸) and (cos Î¸, -sin Î¸)

A (sin Î¸, cos Î¸), B (cos Î¸, -sin Î¸)

AB = √((x2 – x1)2 + (y2 – y1)2

Where, x1 = x coordinate of A = sin Î¸

y1 = y coordinate of A = cos Î¸

x2 = x coordinate of B = cos Î¸

y2 = y coordinates of B = – sin Î¸

= √[(cos Î¸ – sin Î¸)2 + (- sin Î¸ – cos Î¸)2]

= √[cos2 Î¸ + sin2 Î¸ – 2 cos Î¸ sin Î¸ + sin2 Î¸ + cos2 Î¸ + 2 cos Î¸ sin Î¸]

= √2 units

(c) (sec Î¸, tan Î¸) and (-tan Î¸, sec Î¸)

A (sec Î¸, tan Î¸), B (-tan Î¸, sec Î¸)

AB = √((x2 – x1)2 + (y2 – y1)2

Where, x1 = x coordinate of A = sec Î¸

y1 = y coordinate of A = tan Î¸

x2 = x coordinate of B = -tan Î¸

y2 = y coordinates of B = sec Î¸

= √[(- tan Î¸ – sec Î¸)2 + (sec Î¸ – tan Î¸)2]

= √[tan2 Î¸ + sec2 Î¸ – 2 tan Î¸ sec Î¸ + sec2 Î¸ + tan2 Î¸ – 2 tan Î¸ sec Î¸]

= √(2 sec2 Î¸ + 2 tan2 Î¸) units

(d) (sin Î¸ – cosec Î¸, cos Î¸ – cot Î¸) and (cos Î¸ – cosec Î¸, – sin Î¸ – cot Î¸)

A (sin Î¸ – cosec Î¸, cos Î¸ – cot Î¸), B (cos Î¸ – cosec Î¸, – sin Î¸ – cot Î¸)

AB = √((x2 – x1)2 + (y2 – y1)2

Where, x1 = x coordinate of A = sin Î¸ – cosec Î¸

y1 = y coordinate of A = cos Î¸ – cot Î¸

x2 = x coordinate of B = cos Î¸ – cosec Î¸

y2 = y coordinates of B = – sin Î¸ – cot Î¸

= √[((cos Î¸ – cosec Î¸) – (sin Î¸ – cosec Î¸))2 + ((- sin Î¸ – cot Î¸) – (cos Î¸ – cot Î¸))2]

= √[(cos Î¸ – cosec Î¸ – sin Î¸ + cosec Î¸)2 + (- sin Î¸ – cot Î¸ – cos Î¸ + cot Î¸)2]

= √[(cos Î¸ – sin Î¸)2 + (- sin Î¸ – cos Î¸)2]

= √(cos2 Î¸ + sin2 Î¸ – 2 cos Î¸ sin Î¸ + sin2 Î¸ + cos 2 Î¸ + 2 sin Î¸ cos Î¸)

= √2 units

4. Find the distance of a point (7, 5) from another point on the x – axis whose abscissa is – 5.

From the question it is given that, abscissa is – 5.

Let us assume point on x – axis be (x, 0)

So, point P (-5, 0)

Point Q (7, 5)

PQ = √((7 + 5)2 + (5 – 0)2)

= √((12)2 + (5)2)

= √(144 + 25)

= √169

= 13 units

5. Find the distance of a point (13, -9) from another point on the line y = 0 whose abscissa is 1.

From the question it is given that point on the line y = 0 whose abscissa is 1.

So, point is P(1, 0)

Let us assume (13, -9) be point Q.

PQ = √((13 – 1)2 + (-9 – 0)2)

= √((12)2 + (-9)2)

= √(144 + 81)

= √225

= 15 units

6. Find the distance of a point (12, 5) from another point on the line x = 0 whose ordinate is 9.

From the question it is given that, point on the line x = 0 whose ordinate is 9.

So, point is P(0,9)

Let us assume that point (12, 5) be Q.

PQ = √((12 – 0)2 + (5 – 9)2)

= √((12)2 + (-4)2)

= √(144 + 16)

= √160

= 4√10 units

7. Find the value of a if the distance between the points (5, a) and (1, 5) is 5 units.

From the question it is given that,

The distance between the points (5, a) and (1, 5), assume the two points be A and B respectively.

AB = 5 units

√((5 – 1)2 + (a + 5)2) = 5

Now, squaring on both side we get,

(5 – 1)2 + (a + 5)2 = 25

⇒ (25 + 1 – 10) + (a2 + 25 + 10a) = 25

⇒ 16 + a2 + 25 + 10a = 25

⇒ a2 + 10a + 16 = 25 – 25

⇒ a2 + 10a + 16 = 0

⇒ a2 + 8a + 2a + 16 = 0

Now, take out common in above terms we get,

a(a + 8) + 2(a + 8) = 0

⇒ (a + 8) (a + 2) = 0

⇒ a + 8 = 0, a + 2 = 0

⇒ a = -8, a = – 2

8. Find the value of m if the distance between the points (m, -4) and (3, 2) is 3√5 units.

From the question it is given that,

The distance between the points (m, -4) and (3, 2), assume the two points be A and B respectively.

AB = 3√5 units

√((m – 3)2 + (-4 – 2)2) = 3√5

Now, squaring on both side we get,

(m – 3)2 + (-4 – 2)2 = 45

⇒ (m2 + 9 – 6m) + (16 + 4 + 16) = 45

⇒ m2 + 9 – 6m + 36 = 45

By transposing we get,

m2 – 6m + 45 – 45 = 0

⇒ m2 – 6m = 0

Now, take out common in above terms we get,

m(m – 6) = 0

⇒ m = 0, m – 6 = 0

⇒ m = 0, m = 6.

9. Find the relation between a and b if the point P(a, b) is equidistant from A(6, -1) and B(5, 8).

From the question it is given that, point A (6, -1), point B (5, 8)

Then, P(a, b) is equidistant

So, PA = PB

PA2 = PB2

(a – 6)2 + (b + 1)2 = (a – 5)2 + (b – 8)2

We know that,

(a + b)2 = a2 + 2ab + b2 and (a – b)2 = a2 – 2ab + b2

a2 + 36 – 12a + b2 + 1 + 2b = a2 + 25 – 10a + b2 + 64 – 16b

By transposing we get,

a2 – a2 + 37 – 89 – 12a + 10a + b2 – b2 + 2b + 16b = 0

⇒ –52 – 2a + 18b = 0

Divide both side by 2 we get,

-2a/2 + 18b/2 – 52/2 = 0

⇒ -a + 9b – 26 = 0

⇒ a = 9b – 26

10. Find the relation between x and y if the point M(x, y) is equidistant from R(0, 9) and T(14, 11).

From the question it is given that, point R(0, 9) and T(14, 11)

Then, M(x, y) is equidistant from R and T.

So, MR = MT

MR2 = MT2

(x – 0)2 + (y – 9)2 = (x – 14)2 + (y – 11)2

We know that,

(a + b)2 = a2 + 2ab + b2 and (a – b)2 = a2 – 2ab + b2

x2 + y2 + 81 – 18y = x2 + 196 – 28x + y2 + 121 – 22y

By transposing we get,

x2 – x2 + y2 – y2 + 81 – 317 – 18y + 22y + 28x = 0

⇒ – 236 + 4y + 28x = 0

Divide both side by 4 we get,

4y/4 + 28x/4 – 236/4 = 0

⇒ y + 7x – 59 = 0

11. Find the distance between P and Q if P lies on the y – axis and has an ordinate 5 while Q lies on the x – axis and has an abscissa 12.

From the question it is given that,

P lies on the y – axis and has an ordinate 5.

Then, point P (0, 5)

And also in the question it is given that, Q lies on the x – axis and has an abscissa 12.

So, point Q (12, 0)

We know that, PQ = √((x2 – x1)2 + (y2 – y1)2

Where, x1 = x coordinate of P = 0

y1 = y coordinate of Q = 5

x2 = x coordinate of P = 12

y2 = y coordinates of Q = 0

= √((12 – 0)2 + (0 – 5)2)

= √((144) + (25))

= √(169)

= 13 units

12. P and Q are two points lying on the x – axis and the y – axis respectively. Find the coordinates of P and Q if the difference between the abscissa of P and the ordinates of Q is 1 and PQ is 5 units.

From the question it is given that,

P and Q are two points lying on the x – axis and the y – axis respectively.

The difference between the abscissa of P and the ordinates of Q is 1 and PQ is 5 units.

Let us assume abscissa of P be ‘a’ the ordinate of Q is a – 1.

So, P(a, 0), Q(0, a – 1)

Then, √[(x – 0)2 + (0 – x + 1)2] = 5

Now, squaring on both side we get,

x2 + x2 + 1 – 2x = 25

⇒ 2x2 + 1 – 2x = 25

By transposing we get,

2x2 + 1 – 2x – 25 = 0

⇒ 2x2 – 2x – 24 = 0

Divide both side by 4 we get,

2x2/2 – 2x/2 – 24/2 = 0

⇒ x2 – x – 12 = 0

⇒ x2 – 4x + 3x – 12 = 0

Now, take out common in above terms we get,

x(x – 4) + 3(x – 4) = 0

⇒ (x – 4) (x + 3) = 0

⇒ x – 4 = 0, x + 3 = 0

⇒ x = 4, x = -3

Therefore, the coordinates of P are (4, 0) or (-3, 09)

And coordinates of Q are (0, 3) or (0, -4).

13. Find the point on the x – axis equidistant from the points (5, 4) and (-2, 3).

Let us assume the point on x – axis be Q (x, 0) and point A (5, 4), point B (-2, 3)

From the question it is given that, QA = QB

So, QA2 = QB2

(x – 5)2 + (0 – 4)2 = (x + 2)2 + (0 – 3)2

⇒ x2 + 25 – 10x + 16 = x2 + 4 + 4x + 9

On simplification we get,

– 14x + 28 = 0

⇒ 14x = 28

⇒ x = 28/14

⇒ x = 2

Therefore, point on x – axis is (2, 0)

14. A line segment of length 10 units has one end at A (-4, 3). If the ordinate of the other end B is 9, find the abscissa of this end.

From the question it is given that,

Length of line segment i.e. AB = 10 units

Point A is (-4, 3)

Let us assume point B is (y, 9)

Then,

√[(-4 – x)2 + (3 – 9)2] = 10

Now, squaring on both side we get,

(-4 – x)2 + (3 – 9)2 = 100

⇒ 16 + x2 + 8x + 36 = 100

By transposing we get,

16 + 36 – 100 + x2 + 8x = 0

⇒ x2 + 8x – 48

⇒ x2 + 12x – 4x – 48 = 0

Now, take out common in above terms we get,

x(x + 12) – 4(x + 12) = 0

⇒ (x – 4) (x + 12) = 0

⇒ x – 4 = 0, x + 12 = 0

⇒ x = 4, x = 12

Hence, the abscissa of other end is 4 or – 12.

15. Prove that the following sets of points are collinear:

(a) (5, 5), (3, 4), (-7, -1)

(b) (5, 1), (3, 2), (1, 3)

(c) (4, -5), (1, 1), (-2, 7)

Three or more points that lie on the same line are called collinear points.

(a) Let us assume X (5, 5), Y (3, 4) and Z (-7, -1)

Then,

XY = √[(5 – 3)2 + (5 – 4)2]

= √[(2)2 + (1)2]

= √[4 + 1]

= √5 units

YZ = √[(3 + 7)2 + (4 + 1)2]

= √[(10)2 + (5)2]

= √[100 + 25]

= √100

= 5√5 units

XZ = √[(5 + 7)2 + (5 + 1)2]

= √[(12)2 + (6)2]

= √[144 + 36]

= √180

= 6√5 units

XY + YZ = √5 + 5√5

= 6√5

= XZ

So, XY + YZ = XZ

Therefore, X, Y and Z are collinear points.

(b) (5, 1), (3, 2), (1, 3)

Three or more points that lie on the same line are called collinear points.

Let us assume X (5, 1), Y (3, 2) and Z (1, 3)

Then,

XY = √[(5 – 3)2 + (1 – 2)2]

= √[(2)2 + (-1)2]

= √[4 + 1]

= √5 units

YZ = √[(3 – 1)2 + (2 – 3)2]

= √[(2)2 + (-1)2]

= √[4 + 1]

= √5 units

XZ = √[(5 – 1)2 + (1 – 3)2]

= √[(4)2 + (-2)2]

= √[16 + 4]

= √20

= 2√5 units

XY + YZ = √5 + √5

= 2√5

= XZ

So, XY + YZ = XZ

Therefore, X, Y and Z are collinear points.

(c) (4, -5), (1, 1), (-2, 7)

Three or more points that lie on the same line are called collinear points.

Let us assume X (4, -5), Y (1, 1) and Z (-2, 7)

Then,

XY = √[(4 – 1)2 + (-5 – 1)2]

= √[(3)2 + (-6)2]

= √[9 + 36]

= √45

= 3√5 units

YZ = √[(1 + 2)2 + (1 – 7)2]

= √[(3)2 + (6)2]

= √[9 + 36]

= √45

= 3√5

XZ = √[(4 + 2)2 + (-5 – 7)2]

= √[(6)2 + (-12)2]

= √[36 + 144]

= √180

= 6√5 units

XY + YZ = 3√5 + 3√5

= 6√5

= XZ

So, XY + YZ = XZ

Therefore, X, Y and Z are collinear points.

16. Find the coordinates of O, the centre of a circle passing through A (8, 12), B (11, 3) and C (0, 14). Also, find its radius.

17. Find the coordinates of O, the centre of a circle passing through P (3, 0), Q (2, √5) and R (-2 √2,- 1). Also find its radius.

18. Find the coordinates of O, the centre passing through A (-2, - 3), B (-1, 0) and C(7, 6). Also, find its radius.

19. The centre of a circle passing through P(8, 5) is (x + 1, x – 4). Find the coordinates of the centre. If the diameter of the circle is 20 units.

20. A (-2, -3), B (-1, 0) and C(7, - 6) are the vertices of a triangle. Find the circumcentre and the circumradius of the triangle.

21. P (5, - B), Q (2, - 9) and R (2, 1) are the vertices of a triangle. Find the circumcentre and the circumradius of the triangle.

22. X (1, 2) and Y (3, - 4) and Z(5, - 6) are the vertices of a triangle. Find the circumcentre and the circumradius of the triangle.

23. Prove that the points (6, - 1), (5, 8) and (1, 3) are the vertices of an isosceles triangle.

24.  Prove that the points (1, 1), (-4, 4) and (4, 6) are the vertices of an isosceles triangle.

25. Prove that the points (-2, 1), (- 1, 4) and (0, 3) are the vertices of a right-angled triangle.

26. Prove that the points (7, 10), (- 2, 5) and (3, - 4) are vertices of an isosceles right-angled triangle.

27. Prove that the points (1, 1), (- 1, - 1) and (√3, √3) are the vertices of an equilateral triangle.

28. Prove that the points (0, 3), (4, 3) and (2, 3 + 2
√3) are the vertices of an equilateral.

29. Prove that the points (5, 3), (1, 2), (2, - 2) and (6, - 1) are the vertices of a square.

30. Prove that the points (4, 6), (-1, 5), (2, 0) and (3, 1) are the vertices of a rhombus.

31. Prove that the points (0, 0), (3, 2), (7, 7) and (4, 5) are the vertices of a parallelogram.

32. Prove that the points (0, 2), (1, 1), (4, 4) and (3, 5) are the vertices of a rectangle.

33. Prove that the points (a, b), (a + 3, b + 4), (a – 1, b + 7) and (a – 4, b + 3) are the vertices of parallelogram.

34. Prove that the points (0, - 4), (6, 2), (3, 5) and (- 3, - 1) and (- 3, - 1) are the vertices of rectangle.

35. ABCD is a square. If the coordinates of A and C are (5, 4) and (-1, 6); find the coordinates of B and D.

36. PQR is an isosceles triangle. If two of its vertices are P (2, 0) and Q (2, 5), find the coordinates of R. If the length of each of the two equal sides is 3.

37. ABC is an equilateral triangle. If the coordinates of A and B are (1, 1) and (- 1, - 1), find the coordinates of C.

### Exercise 12.2

1. Find the coordinates of a point P which divides the line segment joining

(a) A (3, - 3) and B (6, 9) in the ratio 1 : 2.

(b) M (- 4, - 5) and N (3, 2) in the ratio 2 : 5.

(c) S (2, 6) and R (9, - 8) in the ratio 3 : 4.

(d) D (- 7, 9) and E (15, -2) in the ratio 3 : 4.

(e) A (- 8, - 5) and B (7, 10) in the ratio 2 : 3.

(a)

2. Find the points of trisection of the segment joining A (- 3, 7) and B (3, - 2).

3. Find the coordinates of the points of trisection of the line segment joining the points (3, - 3) and (6, 9).

4. Find the ratio in which the point P (2, 4) divides the line joining points (- 3, 1) and (7, 6).

5. Find the ratio in which the point R (1, 5) divides the line segment joining the points S (- 2, - 1) and T (5, 13).

6. The points A, B and C divides the line segment MN in four equal parts. The coordinates of M and N are (- 1, 10) and (7, - 2) respectively. Find the coordinates of A, B and C.

7. Find the ratio in which the line segment joining A (2, - 3) and B (5, 6) is divide by the x-axis.

8. Find the ratio in which the line segment joining P (4, - 6) and Q (- 3, 8) is divided by the line y = 0.

9. In what ratio is the line joining (2, - 1) and (- 5, 6) divided by the y-axis ?

10. In what ratio is the line joining (2, - 4) and (- 3, 6) divided by the line y = 0 ?

11. Find the ratio in which the line x = 0 divides the join of (- 4, 7) and (3, 0). Also, find the coordinates of the point of intersection.

12. In what ratio does the point (1, a) divided the join of (-1, 4) and (4, - 1)? Also, find the value of a.

13. Find the coordinates of point P which divides line segment joining A (- 3, - 10) and B (3, 2) in such a way that PB : AB = 1.5.

14. Find the ratio in which the line x = - 2 divides the line segment joining (-6, - 1) and (1, 6). Find the coordinates of the point of intersection.

15. Find the ratio in which the line y = - 1 divides the line segment joining (6, 5) and (-2, - 11). Find the coordinates of the point of intersection.

16. The line joining P (- 5, - 6) and Q (3, 2) intersects the y-axis at R. PM and QN are perpendiculars from P and Q on the x-axis. Find the ratio PR : RQ.

17. B is a point on the line segment AC. The coordinates of A and B are (2, 5) and (1, 0). If AC = 3 AB, find the coordinates of C.

18. Q is a point on the line segment AB. The coordinates of A and B are (2, 7) and (7, 12) along the line AB so that AQ = 4BQ. Find the coordinates of Q.

19. The origin O (0, 0), P (- 6, 9) and Q (12, 3) are vertices of triangle OPQ. Point M divides OP in the ratio 1 : 2 and point N divides OQ in the ratio 1 : 2. Find the coordinates of points M and N. Also, show that 3MN = PQ.

20. A (2, 5), B (- 1, 2) and C (5, 8) are the vertices of triangle ABC. Point P and Q lie on AB and AC respectively, such that AP : PB = AQ : QC = 1 : 2. Calculate the coordinates of P and Q. Also, show that 3PQ = 3C.

21. A (30, 20) and B (6, - 4) are two fixed points. Find the coordinates of a point P in AB such that 2PB = AP. Also, find the coordinates of some other point Q in AB such that AB = 6AQ.

22. Show that he line segment joining the points (-3, 10) and (6, - 5) is trisected by the coordinates axis.

23. Show that the lines x = 0 and y = 0 trisect the line segment formed by joining the points (-10, - 4) and (5, 8). Find the points trisection.

### Exercise 12.3

1. Find the midpoints of the line segment joining the following pairs points:

(a) (4, 7) and (10, 15)

(b) (-3, 5) and (9, - 9)

(c) (a + b, b – a) and (a – b, a + b)

(d) (3a – 2b, 5a + 7b) and (a + 4b, a – 3b)

(e) (a + 3, 5b), (3a – 1, 3b + 4)

2. The mid-point of the line segment joining A (- 2, 0) and B (x, y) is P (6, 3). Find the coordinates of B.

3. A lies on the x-axis and B lies on the y-axis. The midpoint of the line segment AB is (4, - 3). Find the coordinate of A and B.

4. P, Q and R are collinear such that PQ = QR. If the coordinates of P, Q and R are (- 5, x), (y, 7), (1, - 3) respectively, find the values of x and y.

5. A, B and C are collinear points such that AB = ½.AC. If the coordinates of A, B and C are (-4, - 4), (-2, b) and (a, 2), find the values of a and b.

6. The midpoint of the line segment joining the points P (2, m) and Q (n, 4) is R (3, 5). Find the values of m and n.

7. The midpoint of the line segment joining the points (p, 2) and (3, 6) is (2, q). Find the numerical values of a and b.

8. The coordinates of the end points of the diameter of a circle are (3, 1) and (7, 11). Find the coordinates of the centre of the circle.

9. AB is a diameter of a circle with centre O. If the coordinates of A and O are (1, 4) and (3, 6). Find the coordinates of B and the length of the diameter.

10. A (6,- 2), B (3, - 2) and C (8, 6) are the three vertices of a parallelogram ABCD. Find the coordinates of the fourth vertex C.

11. P (- 2, 5), Q (3, 6), R (-4, 3) and S (- 9, 2) are the vertices of a quadrilateral. Find the coordinates of the midpoints of the diagonals PR and QS. Give a special name to the quadrilateral.

12. (4, 2) and (-1, 5) are the adjacent vertices of a parallelogram. (-3, 2) are the coordinates of the points of intersection of its diagonal. Find the coordinates of the other two vertices.

13. Three consecutive vertices of a parallelogram ABCD are A (8, 5), B (- 7, - 5) and C (- 5, 5). Find the coordinates of the fourth vertex D.

14. The points (2, - 1), (-1, 4) and (- 2, 2) are midpoints of the sides of a triangle. Find its vertices.

15. The midpoints of the sides of a triangle are (- 2, 3), (4, - 3), (4, 5), find its vertices.

16. If (- 3, 2), (1, - 2) and (5, 6) are the midpoints of the sides of a triangle, find the coordinates of the vertices of the triangle.

17. Find the length of the median through the vertex A of triangle ABC whose vertices A (7, - 3), B (5, 3) and C (3, 1).

18. Find the centroid of a triangle whose vertices are (3, - 5), (- 7, 4) and (10, - 2).

19. Two vertices of a triangle are (1, 4) and (3, 1). If the centroid of the triangle is the origin, find the third vertex.

20. A (4, 2), B (- 2, -6) and C (1, 1) are the vertices of triangle ABC. Find its centroid and the length of the median through C.

21. A triangle is formed by line segments joining the points (5, 1), (3, 4) and (1, 1). Find the coordinates of the centroid.

22. The coordinates of the centroid I of triangle PQR are (2, 5). If Q = (-6, 5) and R = (7, 8). Calculate the coordinates of vertex P.

23. Two vertices of a triangle are (-1, 4) and (5, 2). If the centroid is (0, - 3), find the third vertex.

24. The midpoints of three sides of a triangle are (1, 2), (2, - 3) and (3, 4). Find the centroid of the triangle.

25. ABC is a triangle whose vertices are A (- 4, 2), B (0, 2) and C (-2, - 4). D, E and F are the midpoint of the sides BA, CA and AB respectively. Prove that the centroid of the Î”ABC coincides with the centroid of the Î”DEF.

26. Prove that the points A (-5, 4), B (- 1, - 2) and C (5, 2) are the vertices of an isosceles right-angled triangle. Find the coordinates of D so that ABCD is a square.

27. The centre of a circle is (a + 2, a – 1). Find the value of a, given that the circle passes through the points (2, - 2) and (B, - 2).