# 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)**

**Answer**

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

A = (5, – 2)

B = (1, 5)

AB = √((x_{2} – x_{1})^{2} + (y_{2} – y_{1})^{2}

Where, x_{1} = x coordinate of A = 5

y_{1} = y coordinate of A = – 2

x_{2} = x coordinate of B = 1

y_{2} = 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 = √((x_{2} – x_{1})^{2} + (y_{2} – y_{1})^{2}

Where, x_{1} = x coordinate of A = 1

y_{1} = y coordinate of A = 3

x_{2} = x coordinate of B = 3

y_{2} = 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 = √((x_{2} – x_{1})^{2} + (y_{2} – y_{1})^{2}

Where, x_{1} = x coordinate of A = 7

y_{1} = y coordinate of A = -7

x_{2} = x coordinate of B = 2

y_{2} = 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 = √((x_{2} – x_{1})^{2} + (y_{2} – y_{1})^{2}

Where, x_{1} = x coordinate of A = 4

y_{1} = y coordinate of A = 1

x_{2} = x coordinate of B = -4

y_{2} = 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 = √((x_{2} – x_{1})^{2} + (y_{2} – y_{1})^{2}

Where, x_{1} = x coordinate of A = 13

y_{1} = y coordinate of A = 7

x_{2} = x coordinate of B = 4

y_{2} = 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)**

**Answer**

**(a)** (5, 12)

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

AB = √((x_{2} – x_{1})^{2} + (y_{2} – y_{1})^{2}

Where, x_{1} = x coordinate of A = 0

y_{1} = y coordinate of A = 0

x_{2} = x coordinate of B = 5

y_{2} = 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 = √((x_{2} – x_{1})^{2} + (y_{2} – y_{1})^{2}

Where, x_{1} = x coordinate of A = 0

y_{1} = y coordinate of A = 0

x_{2} = x coordinate of B = 6

y_{2} = 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 = √((x_{2} – x_{1})^{2} + (y_{2} – y_{1})^{2}

Where, x_{1} = x coordinate of A = 0

y_{1} = y coordinate of A = 0

x_{2} = x coordinate of B = 8

y_{2} = 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 = √((x_{2} – x_{1})^{2} + (y_{2} – y_{1})^{2}

Where, x_{1} = x coordinate of A = 0

y_{1} = y coordinate of A = 0

x_{2} = x coordinate of B = 0

y_{2} = 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 = √((x_{2} – x_{1})^{2} + (y_{2} – y_{1})^{2}

Where, x_{1} = x coordinate of A = 0

y_{1} = y coordinate of A = 0

x_{2} = x coordinate of B = 13

y_{2} = 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¸)**

**Answer**

**(a)** (p + q, p – q) and (p – q, p – q)

A (p + q, p – q), B (p – q, p – q)

AB = √((x_{2} – x_{1})^{2} + (y_{2} – y_{1})^{2}

Where, x_{1} = x coordinate of A = p + q

y_{1} = y coordinate of A = p – q

x_{2} = x coordinate of B = p – q

y_{2} = 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 = √((x_{2} – x_{1})^{2} + (y_{2} – y_{1})^{2}

Where, x_{1} = x coordinate of A = sin Î¸

y_{1} = y coordinate of A = cos Î¸

x_{2} = x coordinate of B = cos Î¸

y_{2} = y coordinates of B = – sin Î¸

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

= √[cos^{2} Î¸ + sin^{2} Î¸ – 2 cos Î¸ sin Î¸ + sin^{2} Î¸ + cos^{2} 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 = √((x_{2} – x_{1})^{2} + (y_{2} – y_{1})^{2}

Where, x_{1} = x coordinate of A = sec Î¸

y_{1} = y coordinate of A = tan Î¸

x_{2} = x coordinate of B = -tan Î¸

y_{2} = y coordinates of B = sec Î¸

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

= √[tan^{2} Î¸ + sec^{2} Î¸ – 2 tan Î¸ sec Î¸ + sec^{2} Î¸ + tan^{2} Î¸ – 2 tan Î¸ sec Î¸]

= √(2 sec^{2} Î¸ + 2 tan^{2} 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 = √((x_{2} – x_{1})^{2} + (y_{2} – y_{1})^{2}

Where, x_{1} = x coordinate of A = sin Î¸ – cosec Î¸

y_{1} = y coordinate of A = cos Î¸ – cot Î¸

x_{2} = x coordinate of B = cos Î¸ – cosec Î¸

y_{2} = 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}]

= √(cos^{2} Î¸ + sin^{2} Î¸ – 2 cos Î¸ sin Î¸ + sin^{2} 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.**

**Answer**

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.**

**Answer**

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.**

**Answer**

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.**

**Answer**

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) + (a^{2} + 25 + 10a) = 25

⇒ 16 + a^{2} + 25 + 10a = 25

⇒ a^{2} + 10a + 16 = 25 – 25

⇒ a^{2} + 10a + 16 = 0

⇒ a^{2} + 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.**

**Answer**

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

⇒ (m^{2} + 9 – 6m) + (16 + 4 + 16) = 45

⇒ m^{2} + 9 – 6m + 36 = 45

By transposing we get,

m^{2} – 6m + 45 – 45 = 0

⇒ m^{2} – 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).**

**Answer**

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

Then, P(a, b) is equidistant

So, PA = PB

PA^{2} = PB^{2}

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

We know that,

(a + b)^{2} = a^{2} + 2ab + b^{2} and (a – b)^{2} = a^{2} – 2ab + b^{2}

a^{2} + 36 – 12a + b^{2} + 1 + 2b = a^{2} + 25 – 10a + b^{2} + 64 – 16b

By transposing we get,

a^{2} – a^{2} + 37 – 89 – 12a + 10a + b^{2} – b^{2} + 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).**

**Answer**

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

MR^{2} = MT^{2}

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

We know that,

(a + b)^{2} = a^{2} + 2ab + b^{2} and (a – b)^{2} = a^{2} – 2ab + b^{2}

x^{2} + y^{2} + 81 – 18y = x^{2} + 196 – 28x + y^{2} + 121 – 22y

By transposing we get,

x^{2} – x^{2} + y^{2} – y^{2} + 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.**

**Answer**

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 = √((x_{2} – x_{1})^{2} + (y_{2} – y_{1})^{2}

Where, x_{1} = x coordinate of P = 0

y_{1} = y coordinate of Q = 5

x_{2} = x coordinate of P = 12

y_{2} = 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.**

**Answer**

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,

x^{2} + x^{2} + 1 – 2x = 25

⇒ 2x^{2} + 1 – 2x = 25

By transposing we get,

2x^{2} + 1 – 2x – 25 = 0

⇒ 2x^{2} – 2x – 24 = 0

Divide both side by 4 we get,

2x^{2}/2 – 2x/2 – 24/2 = 0

⇒ x^{2} – x – 12 = 0

⇒ x^{2} – 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).**

**Answer**

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, QA^{2} = QB^{2}

(x – 5)^{2} + (0 – 4)^{2} = (x + 2)^{2} + (0 – 3)^{2}

⇒ x^{2} + 25 – 10x + 16 = x^{2} + 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.**

**Answer**

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 + x^{2} + 8x + 36 = 100

By transposing we get,

16 + 36 – 100 + x^{2} + 8x = 0

⇒ x^{2} + 8x – 48

⇒ x^{2} + 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)**

**Answer**

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.**

**Answer**

**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.**

**Answer**

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

**Answer**

**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. **

**Answer**

**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. **

**Answer**

**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. **

**Answer**

**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. **

**Answer**

**Answer**

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

**Answer**

**Answer**

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

**Answer**

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

**Answer**

**√3) are the vertices of an equilateral.**

**Answer**

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

**Answer**

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

**Answer**

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

**Answer**

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

**Answer**

**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. **

**Answer**

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

**Answer**

**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. **

**Answer**

**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.**

**Answer**

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

**Answer**

**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. **

**Answer**

**(a)**

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

**Answer**

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

**Answer**

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

**Answer**

**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).**

**Answer**

**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. **

**Answer**

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

**Answer**

**Answer**

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

**Answer**

**Answer**

**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. **

**Answer**

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

**Answer**

**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. **

**Answer**

**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. **

**Answer**

**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. **

**Answer**

**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. **

**Answer**

**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. **

**Answer**

**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. **

**Answer**

**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. **

**Answer**

**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. **

**Answer**

**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. **

**Answer**

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

**Answer**

**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. **

**Answer**

**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) **

**Answer**

**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. **

**Answer**

**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. **

**Answer**

**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. **

**Answer**

**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. **

**Answer**

**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. **

**Answer**

**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. **

**Answer**

**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. **

**Answer**

**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. **

**Answer**

**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. **

**Answer**

**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. **

**Answer**

**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. **

**Answer**

**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.**

**Answer**

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

**Answer**

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

**Answer**

**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.**

**Answer**

**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). **

**Answer**

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

**Answer**

**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.**

**Answer**

**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. **

**Answer**

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

**Answer**

**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. **

**Answer**

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

**Answer**

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

**Answer**

**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. **

**Answer**

**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. **

**Answer**

**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). **

**Answer**

**28. Let A (- a, 0), B (0, a) and C (Î±, Î²) be the vertices of the Î”ABC and G be its centroid. Prove that **

**GA ^{2} + GB^{2} + GC^{2} = 1/3(AB^{2} + BC^{2} + CA^{2}). **

**Answer**

**29. A (2, 5), B (- 2, 4) and C (- 2, 6) are the vertices of a vertices of a triangle ABC. Prove that ABC is an isosceles triangle. **

**Answer**