# ICSE Solutions for Selina Concise Chapter 13 Section and Mid-point Formula Class 10 Maths

### Exercise 13(A)

1. Calculate the co-ordinates of the point P which divides the line segment joining:

(i) A (1, 3) and B (5, 9) in the ratio 1: 2.

(ii) A (-4, 6) and B (3, -5) in the ratio 3: 2.

Solution

(i) Let’s assume the co-ordinates of the point P be (x, y)

Then by section formula, we have

P(x, y) = (m1x2 + m2x1)/ (m1 + m2), (m1y2 + m2y1)/ (m1 + m2)

Hence, the co-ordinates of point P are (7/3, 5).

(ii) Let’s assume the co-ordinates of the point P be (x, y)

Then by section formula, we have

P(x, y) = (m1x2 + m2x1)/(m1 + m2), (m1y2 + m2y1)/(m1 + m2)

Hence, the co-ordinates of point P are (1/5, -3/5).

2. In what ratio is the line joining (2, -3) and (5, 6) divided by the x-axis.

Solution

Let’s assume the joining points as A(2, -3) and B(5, 6) be divided by point P(x ,0) in the ratio k: 1.

Then we have,

y = ky2 + y1/ (k + 1)

⇒ 0 = 6k + (-3)/(k + 1)

⇒ 0 = 6k – 3

⇒ k = ½

Hence, the required ratio is 1: 2.

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

Solution

Let’s assume the line joining points A(2, -4) and B(-3, 6) be divided by point P (0, y) in the ratio k: 1.

Then we have,

x = kx2 + x1/ (k + 1)

⇒ 0 = k(-3) + (1×2)/ (k + 1)

⇒ 0 = -3k + 2

⇒ k = 2/3

Hence, the required ratio is 2: 3.

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

Solution

Let’s assume the point P (1, a) divide the line segment AB in the ratio k: 1.

Then by section formula, we have

1 = (4k – 1)/ (k + 1),

⇒ k + 1 = 4k – 1

⇒ 2 = 3k

⇒ k = 2/3 ….. (1)

And,

a = (-k + 4)/ (k + 1)

⇒ a = (-(2/3) + 4)/(2/3 + 1) [using (1)]

⇒ a = 10/5 = 2

Thus, the required ratio is 2: 3 and a = 2.

5. In what ratio does the point (a, 6) divide the join of (-4, 3) and (2, 8)? Also, find the value of a.

Solution

Let’s assume the point P (a, 6) divides the line segment joining A (-4, 3) and B (2, 8) in the ratio k: 1.

Then by section formula, we have

6 = (8k + 3)/ (k + 1),

⇒ 6k + 6 = 8k + 3

⇒ 3 = 3k

⇒ k = 3/2 …. (1)

a = (2k – 4)/ (k + 1)

⇒ a = (2(3/2) – 4)/(3/2 + 1)

⇒ a = -2/5

Thus, the required ratio is 3: 2 and a = -2/5

6. In what ratio is the join of (4, 3) and (2, -6) divided by the x-axis. Also, find the co-ordinates of the point of intersection.

Solution

Let’s assume the point P (x, 0) on x-axis divides the line segment joining A (4, 3) and B (2, -6) in the ratio k: 1.

Then by section formula, we have

0 = (-6k + 3)/ (k + 1)

⇒ 0 = -6k + 3

⇒ k = ½

Hence, the required ratio is 1: 2

And,

x = (2k + 4)/(k + 1)

= {2(1/2) + 4}/(k + 1)

= 10/3

Therefore, the required co-ordinates of the point of intersection are (10/3, 0).

7. Find the ratio in which the join of (-4, 7) and (3, 0) is divided by the y-axis. Also, find the coordinates of the point of intersection.

Solution

Let’s assume S (0, y) be the point on y-axis which divides the line segment PQ in the ratio k: 1.

Then by section formula, we have

0 = (3k – 4)/(k + 1)

⇒ 3k = 4

⇒ k = 4/3 … (1)

y = (0 + 7)/ (k + 1)

⇒ y = 7/(4/3 + 1) [From (1)]

⇒ y = 3

Thus, the required ratio is 4: 3 and the required point is S(0, 3).

8. Points A, B, C and D divide the line segment joining the point (5, -10) and the origin in five equal parts. Find the co-ordinates of A, B, C and D.

Solution

9. The line joining the points A (-3, -10) and B (-2, 6) is divided by the point P such that PB/AB = 1/5 Find the co-ordinates of P.

Solution

Let the coordinates of point P be taken as (x, y).

Given,

PB: AB = 1: 5

So, PB: PA = 1: 4

Hence, the coordinates of P are

10. P is a point on the line joining A (4, 3) and B (-2, 6) such that 5AP = 2BP. Find the co-ordinates of P.

Solution

5AP = 2BP

So, AP/BP = 2/5

Hence, the co-ordinates of the point P are

((2×(-2) + 5×4)/(2 + 5), (2×6 + 5×3)/ (2 + 5))

= (16/7, 27/7)

11. Calculate the ratio in which the line joining the points (-3, -1) and (5, 7) is divided by the line x = 2. Also, find the co-ordinates of the point of intersection.

Solution

We know that,

The co-ordinates of every point on the line x = 2 will be of the type (2, y).

So from section formula, we have

x = m1×5 + m2×(-3)/(m1 + m2)

⇒ 2 = 5m1 – 3m2/ m1 + m2

⇒ 2m1 + 2m2 = 5m1 – 3m2

⇒ 5m2 = 3m1

Hence, the required ratio is 5: 3.

y = (m1×7 + m2×(-1))/(m1 + m2)

⇒ y = 5×7 + 3(-1)/ 5 + 3

⇒ y = 35 – 3/ 8

⇒ y = 32/8 = 4

Therefore, the required co-ordinates of the point of intersection are (2, 4).

12. Calculate the ratio in which the line joining A (6, 5) and B (4, -3) is divided by the line y = 2.

Solution

We know that,

The co-ordinates of every point on the line y = 2 will be of the type (x, 2).

So, by section formula, we have

⇒ 2m1 + 2m2 = -3m1 + 5m2

⇒ 5m1 = 3m2

⇒ m1/ m2 = 3/5

Hence, the required ratio is 3: 5.

13. The point P (5, -4) divides the line segment AB, as shown in the figure, in the ratio 2: 5. Find the co-ordinates of points A and B. Given AP is smaller than BP

Solution

From the diagram we can see that,

Point A lies on x-axis. So, its co-ordinates can be taken as A (x, 0).

Point B lies on y-axis. So, its co-ordinates can be taken as B be (0, y).

And, P divides AB in the ratio 2: 5. (Given)

Now, we have

⇒ 5 = 5x/7

⇒ x = 7

Hence, the co-ordinates of point A are (7, 0).

-4 = 2y/7

-2 = y/7

y = -14

Hence, the co-ordinates of point B are (0, -14)

### Exercise 13(B)

1. Find the mid-point of the line segment joining the points:

(i) (-6, 7) and (3, 5)

(ii) (5, -3) and (-1, 7)

Solution

(i) Let A (-6, 7) and B (3, 5)

So, the mid-point of AB = (-6+3/2, 7+5/2) = (-3/2, 6)

(ii) Let A (5, -3) and B (-1, 7)

So, the mid-point of AB = (5-1/2, -3+7/2) = (2, 2)

2. Points A and B have co-ordinates (3, 5) and (x, y) respectively. The mid-point of AB is (2, 3). Find the values of x and y.

Solution

Given, mid-point of AB = (2, 3)

Thus,

(3+ x/2, 5+ y/2) = (2, 3)

⇒ 3 + x/2 = 2 and 5 + y/2 = 3

⇒ 3 + x = 4 and 5 + y = 6

⇒ x = 1 and y = 1

3. A (5, 3), B (-1, 1) and C (7, -3) are the vertices of triangle ABC. If L is the mid-point of AB and M is the mid-point of AC, show that LM = ½ BC.

Solution

It’s given that, L is the mid-point of AB and M is the mid-point of AC.

Co-ordinates of L are,

Co-ordinates of M are,

Using distance formula, we have:

Thus, LM = ½ BC

4. Given M is the mid-point of AB, find the co-ordinates of:

(i) A; if M = (1, 7) and B = (-5, 10)

(ii) B; if A = (3, -1) and M = (-1, 3).

Solution

(i) Let’s assume the co-ordinates of A to be (x, y).

So, (1, 7) = (x-5/2, y+10/2)

1 = x- 5/2 and 7 = y+ 10/2

⇒ 2 = x – 5 and 14 = y + 10

⇒ x = 7 and y = 4

Thus, the co-ordinates of A are (7, 4).

(ii) Let’s assume the co-ordinates of B be (x, y).

So, (-1, 3) = (3+x/2, -1+y/2)

-1 = 3+x/2 and 3 = -1+y/2

⇒ -2 = 3 + x and 6 = -1 + y

⇒ x = -5 and y = 7

Thus, the co-ordinates of B are (-5, 7).

5. P (-3, 2) is the mid-point of line segment AB as shown in the given figure. Find the co-ordinates of points A and B.

Solution

It’s seen that,

Point A lies on y-axis, hence its co-ordinates is taken to be (0, y).

Point B lies on x-axis, hence its co-ordinates is taken to be (x, 0).

Given, P (-3, 2) is the mid-point of line segment AB.

So, by the mid-point section we have

(-3, 2) = (0+x/2, y+0/2)

⇒ (-3, 2) = (x/2, y/2)

⇒ -3 = x/2 and 2 = y/2

⇒ x = -6 and y = 4

Therefore, the co-ordinates of points A and B are (0, 4) and (-6, 0) respectively.

6. In the given figure, P (4, 2) is mid-point of line segment AB. Find the co-ordinates of A and B.

Solution

Let point A lies on x-axis, hence its co-ordinates can be (x, 0).

And, Point B lies on y-axis, hence its co-ordinates can be (0, y).

Given, P (4, 2) is the mid-point of line segment AB.

So,

(4, 2) = (x+ 0/2, 0+ y/2)

⇒ 4 = x/2 and 2 = y/2

⇒ 8 = x and 4 = y

Thus, the co-ordinates of points A and B are (8, 0) and (0, 4) respectively.

7. (-5, 2), (3, -6) and (7, 4) are the vertices of a triangle. Find the length of its median through the vertex (3, -6).
Solution

Let A (-5, 2), B (3, -6) and C (7, 4) be the vertices of the given triangle.

And let AD be the median through A, BE be the median through B and CF be the median through C.

We know that, median of a triangle bisects the opposite side.

So, the co-ordinates of point F are

The co-ordinates of point D are

And, co-ordinates of point E are

The median of the triangle through the vertex B (3, -6) is BE.

Therefore, by distance formula we get

8. Given a line ABCD in which AB = BC = CD, B = (0, 3) and C = (1, 8). Find the co-ordinates of A and D.

Solution

Given, AB = BC = CD

So, B is the mid-point of AC. Let the co-ordinates of point A be (x, y).

(0, 3) = (x+1/2, y+8/2)

⇒ 0 = (x+1)/2 and 3 = (y+8)/2

⇒ 0 = x + 1 and 6 = y + 8

⇒ x = -1 and y = -2

Hence, the co-ordinates of point A are (-1, -2).

Also given, C is the mid-point of BD. And, let the co-ordinates of point D be (p, q).

(1, 8) = (0+ p/2, 3+ q/2)

⇒ 1= 0+ p/2 and 8 = 3+ q/2

⇒ 2 = 0 + p and 16 = 3 + q

⇒ p = 2 and q = 13

Hence, the co-ordinates of point D are (2, 13).

9. One end of the diameter of a circle is (-2, 5). Find the co-ordinates of the other end of it, if the centre of the circle is (2, -1).

Solution

We know that,

The centre is the mid-point of any diameter of a circle.

Let assume the required co-ordinates of the other end of mid-point to be (x, y).

⇒ 2 = (-2 + x)/2 and -1 = (5 + y)/2

⇒ 4 = -2 + x and -2 = 5 + y

⇒ x = 6 and y = -7

Therefore, the required co-ordinates of the other end of the diameter are (6, -7).

### Exercise 13(C)

1. Given a triangle ABC in which A = (4, -4), B = (0, 5) and C = (5, 10). A point P lies on BC such that BP: PC = 3: 2. Find the length of line segment AP.

Solution

Given, BP: PC = 3: 2

Then by section formula, the co-ordinates of point P are given as:

= (15/5, 40/5)

= (3, 8)

Now, by using distance formula, we get

2. A (20, 0) and B (10, -20) are two fixed points. Find the co-ordinates of a point P in AB such that: 3PB = AB. Also, find the co-ordinates of some other point Q in AB such that AB = 6AQ.

Solution

Given, 3PB = AB

So,

AB/PB = 3/1

(AB – PB)/ PB = (3 – 1)/ 1

AP/PB = 2/1

By section formula, we get the coordinates of P to be

= P (40/3, -40/3)

Also given that, AB = 6AQ

AQ/AB = 1/6

⇒ AQ/(AB – AQ) = 1/(6 – 1)

⇒ AQ/ QB = 1/5

Now, again by using section formula we get

The coordinates of Q as

= Q(110/6, -20/6)

= Q(55/3, -10/3)

3. A (-8, 0), B (0, 16) and C (0, 0) are the vertices of a triangle ABC. Point P lies on AB and Q lies on AC such that AP: PB = 3: 5 and AQ: QC = 3: 5. Show that: PQ = 3/8 BC.

Solution

Given that, point P lies on AB such that AP: PB = 3: 5.

So, the co-ordinates of point P are given as

= (-40/8, 48/8)

= (-5, 6)

Also given that, point Q lies on AB such that AQ: QC = 3: 5.

So, the co-ordinates of point Q are given as

= (-40/8, 0/8)

= (-5, 0)

Now, by distance formula we get

Thus,

3/8 ×BC = 3/8 ×16 = 6 = PQ

Hence, proved.

4. Find the co-ordinates of points of trisection of the line segment joining the point (6, -9) and the origin.

Solution

Let’s assume P and Q to be the points of trisection of the line segment joining A (6, -9) and B (0, 0).

So, P divides AB in the ratio 1: 2.

Hence, the co-ordinates of point P are given as

= (12/3, -18/3)

= (4, -6)

And, Q divides AB in the ratio 2: 1.

Hence, the co-ordinates of point Q are

= (6/3, -9/3) = (2, 3)

Therefore, the required coordinates of trisection of PQ are (4, -6) and (2, -3).

5. A line segment joining A (-1, 5/3) and B (a, 5) is divided in the ratio 1: 3 at P, point where the line segment AB intersects the y-axis.

(i) Calculate the value of ‘a’.

(ii) Calculate the co-ordinates of ‘P’.

Solution

As, the line segment AB intersects the y-axis at point P, let the co-ordinates of point P be taken as (0, y).

And, P divides AB in the ratio 1: 3.

So,

(0, y) = (a-3/4, 10/4)

⇒ 0 = a- 3/4 and y = 10/4

⇒ a -3 = 0 and y = 5/2

⇒ a = 3

Therefore, the value of a is 3 and the co-ordinates of point P are (0, 5/2).

6. In what ratio is the line joining A (0, 3) and B (4, -1) divided by the x-axis? Write the co-ordinates of the point where AB intersects the x-axis.

Solution:

Let assume that the line segment AB intersects the x-axis by point P (x, 0) in the ratio k: 1.

⇒ 0 = (-k + 3)/ (k + 1)

⇒ k = 3

Therefore, the required ratio in which P divides AB is 3: 1.

Also,

x = 4k/(k + 1)

⇒ x = (4×3)/ (3 + 1)

⇒ x = 12/3 = 3

Hence, the co-ordinates of point P are (3, 0).

7. The mid-point of the segment AB, as shown in diagram, is C (4, -3). Write down the co-ordinates of A and B.

Solution

As, point A lies on x-axis, we can assume the co-ordinates of point A to be (x, 0).

As, point B lies on y-axis, we can assume the co-ordinates of point B to be (0, y).

And given, the mid-point of AB is C (4, -3).

⇒ (4, -3) = (x/2, y/2)

⇒ 4 = x/2 and -3 = y/2

⇒ x = 8 and y = -6

Therefore, the co-ordinates of point A are (8, 0) and the co-ordinates of point B are (0, -6).

8. AB is a diameter of a circle with centre C = (-2, 5). If A = (3, -7), find

(i) the length of radius AC

(ii) the coordinates of B.

Solution

(i) Radius AC = √[ (3 + 2)2 + (-7 – 5)2 ]

= √[(52 + (-12)2]

= √(25 + 144)

= √169 = 13 units

(ii) Let the coordinates of B be (x, y).

Now, by mid-point formula, we get

-2 = (3 + x)/2 and 5 = (-7 + y)/2

⇒ -4 = 3 + x and 10 = -7 + y

⇒ x = -7 and y = 17

Hence, the coordinates of B are (-7, 17).

9. Find the co-ordinates of the centroid of a triangle ABC whose vertices are:

A (-1, 3), B (1, -1) and C (5, 1)

Solution

By the centroid of a triangle formula, we get

The co- ordinates of the centroid of triangle ABC as

= (5/3, 1)

10. The mid-point of the line-segment joining (4a, 2b – 3) and (-4, 3b) is (2, -2a). Find the values of a and b.

Solution

Given that the mid-point of the line-segment joining (4a, 2b – 3) and (-4, 3b) is (2, -2a).

So, we have

2 = (4a – 4)/2

⇒ 4 = 4a – 4

⇒ 8 = 4a

⇒ a = 2

Also,

-2a = (2b – 3 + 3b)/2

⇒ -2 ×2 = (5b – 3)/2

⇒ -8 = 5b – 3

⇒ -5 = 5b

⇒ b = -1

11. The mid-point of the line segment joining (2a, 4) and (-2, 2b) is (1, 2a + 1). Find the value of a and b.

Solution

Given,

The mid-point of (2a, 4) and (-2, 2b) is (1, 2a + 1)

So, by using mid-point formula, we know

(1, 2a + 1) = (2a – 2/2, 4 + 2b/2)

⇒ 1 = 2a – 2/2 and 2a + 1 = 4 + 2b/2

⇒ 2 = 2a – 2 and 4a + 2 = 4 + 2b

⇒ 4 = 2a and 4a = 2 + 2b

⇒ a = 2 and 4(2) = 2 + 2b [using the value of a]

⇒ 8 = 2 + 2b

⇒ 6 = 2b

⇒ b = 3

Hence, the value of a = 2 and b = 3.