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# Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 13 - Section and Mid-Point Formula

## Selina Textbook Solutions Chapter 13 - Section and Mid-Point Formula

Selina Textbook Solutions are a perfect way to ace your examination with high marks. These Textbook Solutions are extremely helpful for solving difficult questions in the ICSE Class 10 Mathematics exam. Our Selina Textbook Solutions are written by our subject experts. Find all the answers to the Selina textbook questions of Chapter 13 - Section and Mid-Point Formula.

All Selina textbook questions of Chapter 13 - Section and Mid-Point Formula solutions are created in accordance with the latest ICSE syllabus. These free Textbook Solutions for ICSE Class 10 Selina Concise Mathematics will give you a deeper insight on the fundamentals in this chapter and will help you to score more marks in the final examination. ICSE Class 10 students can refer to these solutions while doing their homework and while studying and revising for the Mathematics exam.

Exercise/Page

## Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 13 - Section and Mid-Point Formula Page/Excercise 13(A)

Solution 1

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

Thus, the co-ordinates of point P are.

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

Thus, the co-ordinates of point P are.

Solution 2

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

Thus, the required ratio is 1: 2.

Solution 3

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

Thus, the required ratio is 2: 3.

Solution 4

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

Using section formula, we have:

Solution 5

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 6

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 7

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 8

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 9

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

Solution 10

5AP = 2BP

The co-ordinates of the point P are

Solution 11

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 12

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 13

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 14

Let P and Q be the point 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 15

Let P and Q be the point 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 16

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 17

(i) A (-4,3) and B (8, -6)

AB =

(ii) Let P be the point, which divides AB on the x-axis in the ratio k : 1.

Therefore, y-co-ordinate of P = 0.

= 0

-6k + 3 = 0

k =

Required ratio is 1: 2.

Solution 18

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 19

(i) Co-ordinates of P are

Co-ordinates of Q are

(ii) Using distance formula, we have:

BC =

PQ =

Hence, PQ = BC.

Solution 20

BP: PC = 2: 3

Co-ordinates of P are

Using distance formula, we have:

Solution 21

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 22

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 23

(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 24

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.

Solution 25

(i)

(ii)

(iii)

## Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 13 - Section and Mid-Point Formula Page/Excercise 13(B)

Solution 1

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

Mid-point of AB =

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

Mid-point of AB =

Solution 2

Mid-point of AB = (2, 3)

Solution 3

Given, 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:

Solution 4

(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 5

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 6

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

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

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

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

Solution 7

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 8

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 9

We know that the centre is the mid-point of diameter.

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

Thus, the required co-ordinates are (6, -7).

Solution 10

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 11

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 12

Let the co-ordinates of vertex C be (x, y).

ABCD is a parallelogram.

Mid-point of AC = Mid-point of BD

Thus, the co-ordinates of vertex C is (5, 8).

Solution 13

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 14

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

Solution 15

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 16

Co-ordinates of the centroid of triangle ABC are

Solution 17

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 18

Given, centroid of triangle ABC is the origin.

## Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 13 - Section and Mid-Point Formula Page/Excercise 13(C)

Solution 1

Given, BP: PC = 3: 2

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

Using distance formula, we have:

Solution 2

Using section formula,

Given, AB = 6AQ

Using section formula,

Solution 3

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 AB such that AQ: QC = 3: 5.

The co-ordinates of point Q are

Using distance formula,

Hence, proved.

Solution 4

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

P divides AB in the ratio 1: 2. Therefore, the co-ordinates of point P are

Q divides AB in the ratio 2: 1. Therefore, the co-ordinates of point Q are

Thus, the required points are (4, -6) and (2, -3).

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 by 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 8

Solution 9

Co- ordinates of the centroid of triangle ABC are

Solution 10

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

Solution 11

Mid-point of (2a, 4) and (-2, 2b) is (1, 2a + 1), therefore using mid-point formula, we have:

y =

2a + 1 =

a = 2

Putting, a = 2 in 2a + 1 = 2 + b, we get,

5 - 2 = b b = 3

Therefore, a = 2, b = 3.

Solution 12

(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 13

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 14

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 15

Solution 16

Solution 17

Solution 18

Solution 19

Solution 20

Solution 21

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