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# Class 10 SELINA Solutions Maths Chapter 13 - Section and Mid-Point Formula

## Section and Mid-Point Formula Exercise TEST YOURSELF

### Solution 1(a)

Correct option: (iv) A = (8, 0) and B = (0, 6)

As A lies on x-axis and P = (4, 3) is the mid-point of AB.

A = (8, 0)

As B lies on y-axis and P = (4, 3) is the mid-point of AB.

B = (0, 6)

### Solution 1(b)

Correct option: (iii) x = –4, y = –8

Mid-point of AB = (0, 0)

x = 4 and y = 8

### Solution 1(c)

Correct option: (ii) (–2, 10)

Let the coordinates of the third vertex be (x, y).

x = 2 and y = 10

Thus, coordinates of the third vertex are (–2, 10).

### Solution 1(d)

Correct option: (iv) (0, 8)

As A, B, C and D divided the join of O and P into five equal parts.

D divides the join of O and P in the ratio 4 : 1.

Let D = (x, y)

(0, 8) = (x, y)

Thus, D = (0, 8).

### Solution 1(e)

Correct option: (ii) 3 : 2

Le the required ratio be k : 1.

Hence, the required ratio is 3 : 2.

### Solution 2

Given, BP: PC = 3: 2

Using section formula, the co-ordinates of point P are

Using distance formula, we have:

### Solution 3

Using section formula,

Given, AB = 6AQ

Using section formula,

### Solution 4

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

The co-ordinates of point P are

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

The co-ordinates of point Q are

Using distance formula,

Hence, proved.

### Solution 5

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

P divides AB in the ratio 1: 3.

Thus, the value of a is 3 and the co-ordinates of point P are.

### Solution 6

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

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

Also, we have:

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

### Solution 7

Since, point A lies on x-axis, let the co-ordinates of point A be (x, 0).

Since, point B lies on y-axis, let the co-ordinates of point B be (0, y).

Given, mid-point of AB is C (4, -3).

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

### Solution 9

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

### Solution 10

(i) Co-ordinates of point P are

(ii) OP =

(iii) Let AB be divided by the point P (0, y) lying on y-axis in the ratio k: 1.

Thus, the ratio in which the y-axis divide the line AB is 4: 17.

### Solution 11

We have:

AB = BC and

ABC is an isosceles right-angled triangle.

Let the coordinates of D be (x, y).

If ABCD is a square, then,

Mid-point of AC = Mid-point of BD

x = 1, y = 8

Thus, the co-ordinates of point D are (1, 8).

### Solution 12

Given, M is the mid-point of the line segment joining the points A (-3, 7) and B (9, -1).

The co-ordinates of point M are

Also, given that, R (2, 2) divides the line segment joining M and the origin in the ratio p: q.

Thus, the ratio p: q is 1: 2.

### Solution 16

Let A' = (x, y) be the image of the point A(5, -3), under reflection in the point P(-1, 3).

P(-1, 3) is the mid - point of the line segment AA'.

Therefore the image of the point A(5, -3), under reflection in the point P(-1, 3) is A'(-7, 9).

### Solution 18

Centroid of a ABC = …… (i)

G(3, 4) is a centroid of ABC ….. given

Therefore, the coordinates of B and C are (5, 4) and (1, 7) respectively.

Length of the side BC

## Section and Mid-Point Formula Exercise Ex. 13(A)

### Solution 2

Let the point P (1, a) divides the line segment AB in the ratio k: 1.

Using section formula, we have:

### Solution 3

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

Using section formula, we have:

### Solution 4

Let 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.

Using section formula, we have:

Thus, the required ratio is 1: 2.

Also, we have:

Thus, the required co-ordinates of the point of intersection are .

### Solution 5

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

Using section formula, we have:

### Solution 6

Point A divides PO in the ratio 1: 4.

Co-ordinates of point A are:

Point B divides PO in the ratio 2: 3.

Co-ordinates of point B are:

Point C divides PO in the ratio 3: 2.

Co-ordinates of point C are:

Point D divides PO in the ratio 4: 1.

Co-ordinates of point D are:

### Solution 7

Let the co-ordinates of point P are (x, y).

### Solution 8

5AP = 2BP

The co-ordinates of the point P are

### Solution 9

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

Using section formula, we have:

Thus, the required ratio is 5: 3.

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

### Solution 10

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

Using section formula, we have:

Thus, the required ratio is 3: 5.

### Solution 11

Point A lies on x-axis. So, let the co-ordinates of A be (x, 0).

Point B lies on y-axis. So, let the co-ordinates of B be (0, y).

P divides AB in the ratio 2: 5.

We have:

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

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

### Solution 12

Let P and Q be the points of trisection of the line segment joining the points A(-3, 0) and B(6, 6).

So, AP = PQ = QB

We have AP: PB = 1: 2

Co-ordinates of the point P are

We have AQ: QB = 2: 1

Co-ordinates of the point Q are

### Solution 13

Let P and Q be the points of trisection of the line segment joining the points A (-5, 8) and B (10, -4).

So, AP = PQ = QB

We have AP: PB = 1: 2

Co-ordinates of the point P are

We have AQ: QB = 2: 1

Co-ordinates of the point Q are

So, point Q lies on the x-axis.

Hence, the line segment joining the given points A and B is trisected by the co-ordinate axes.

### Solution 14

Let A and B be the point of trisection of the line segment joining the points P (2, 1) and Q (5, -8).

So, PA = AB = BQ

We have PA: AQ = 1: 2

Co-ordinates of the point A are

Hence, A (3, -2) is a point of trisection of PQ.

We have PB: BQ = 2: 1

Co-ordinates of the point B are

### Solution 15

Since, point L lies on y-axis, its abscissa is 0.

Let the co-ordinates of point L be (0, y). Let L divides MN in the ratio k: 1.

Using section formula, we have:

Thus, the required ratio is 5: 3.

### Solution 16

BP: PC = 2: 3

Co-ordinates of P are

Using distance formula, we have:

### Solution 17

Since, point K lies on x-axis, its ordinate is 0.

Let the point K (x, 0) divides AB in the ratio k: 1.

Thus, K divides AB in the ratio 3: 5.

Also, we have:

Thus, the co-ordinates of the point K are .

### Solution 18

Since, point K lies on y-axis, its abscissa is 0.

Let the point K (0, y) divides AB in the ratio k: 1.

Thus, K divides AB in the ratio 2: 3.

Also, we have:

Thus, the co-ordinates of the point K are .

### Solution 19

(i) Let point R (0, y) divides PQ in the ratio k: 1.

We have:

Thus, PR: RQ = 4: 3

(ii) Also, we have:

Thus, the co-ordinates of point R are .

= (PM + QN) MN

= (5 + 2) 7

= 7 7

= 24.5 sq units

### Solution 20

Given, A lies on x-axis and B lies on y-axis.

Let the co-ordinates of A and B be (x, 0) and (0, y) respectively.

Given, P is the point (-4, 2) and AP: PB = 1: 2.

Using section formula, we have:

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

(i)

(ii)

(iii)

### Solution 22

Take (x1 , y1) = (-3, 3a + 1) ; (x2 , y2) = (5, 8a) and

(x, y) = (-b, 9a - 2)

Here m1 = 3 and m2 =1

### Solution 1(a)

Correct option: (i)

Let the coordinates of point P are (x, y).

Thus,

### Solution 1(b)

Correct option: (iv) 1 : 3

Let the ratio AP : PB be m1 : m2.

Thus, AP : PB = 1 : 3.

### Solution 1(c)

Correct option: (ii) 5 : 3

Let the required ratio be k : 1 and the point on the line x = 7 is (7, y).

Thus, the line x = 7 divides the join of (2, 4) and (10, 12) is the ratio 5:3.

### Solution 1(d)

Correct option: (i) 3 : 5

Let the required ratio be k : 1 and the point on the line y = 4 is (x, 4).

### Solution 1(e)

Correct option: (ii) 2 : 5

Let the required ratio be k : 1 and the point on y-axis is (0, y).

Hence, the required ratio is 2 : 5.

## Section and Mid-Point Formula Exercise Ex. 13(B)

### Solution 2

Mid-point of AB = (2, 3)

### Solution 3

(i) Let the co-ordinates of A be (x, y).

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

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

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

### Solution 4

Point A lies on y-axis, so let its co-ordinates be (0, y).

Point B lies on x-axis, so let its co-ordinates be (x, 0).

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

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

### Solution 5

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

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.

Co-ordinates of point F are

Co-ordinates of point D are

Co-ordinates of point E are

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

Using distance formula,

### Solution 6

Given, AB = BC = CD

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

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

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

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

### Solution 7

Co-ordinates of the mid-point of AC are

Co-ordinates of the mid-point of BD are

Since, mid-point of AC = mid-point of BD

Hence, ABCD is a parallelogram.

### Solution 8

Let the coordinates of R and S be (x,y) and (a,b) respectively.

Mid-point of PR is O.

O(-3,2) =

-6 = 4 + x, 4 = 2 + y

x = -10 , y = 2

Hence, R = (-10,2)

Similarly, the mid-point of SQ is O.

Thus, the coordinates of the point R and S are (-10, 2) and (-5, -1).

### Solution 9

Let A(x1,y1), B and C be the co-ordinates of the vertices of ABC.

Midpoint of AB, i.e. D

Similarly,

Adding (1), (3) and (5), we get,

From (3)

From (5)

Adding (2), (4) and (6), we get,

From (4)

From (6)

Thus, the co-ordinates of the vertices of ABC are (3, 1), (1, -3) and (-5, 7).

### Solution 10

Given, AB = BC, i.e., B is the mid-point of AC.

### Solution 11

Given, PR = 2QR

Now, Q lies between P and R, so, PR = PQ + QR

PQ + QR = 2QR

PQ = QR

Q is the mid-point of PR.

### Solution 12

Let G be the centroid of DPQR whose coordinates are (2, -5) and let (x,y) be the coordinates of vertex P.

Coordinates of G are,

6 = x + 5, -15 = y + 13

x = 1, y = -28

Coordinates of vertex P are (1, -28)

### Solution 13

Given, centroid of triangle ABC is the origin.

### Solution 1(a)

Correct option: (iii) (–1, 0)

Let P = (7, 8) and Q = (x, y) be the other end of the diameter.

As A is the centre of the circle.

7 + x = 6, 8 + y = 8

x = 1, y = 0

Thus, Q = (–1, 0).

### Solution 1(b)

Correct option: (i) (4, 0)

Coordinates of A are (x, 0)

Coordinates of B are (0, y)

As P bisects the line segment AB, we have

x = 4, y = 4

Thus, Q = (4, 0)

### Solution 1(c)

Correct option: (ii) (–2, 0)

Let D = (x, y).

As ABCD is a parallelogram, we have

Mid-point of AC = Mid-point of BD

x = 2 and y = 0

Thus, D = (–2, 0).

### Solution 1(d)

Correct option: (iv) (–2, 13)

When point P(2, –7) is reflected in the point (0, 3), the coordinates of its image are (–2, 13).

### Solution 1(e)

Correct option: (iii) (2, –3)

Coordinates of the centroid of the triangle is given by: