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

The Indian Certificate of Secondary Education Board emphasises academic excellence, holistic development, and practical application of student's knowledge, and hence it is recognised globally. The board aims to provide a well-rounded education with a curriculum that helps students develop a strong foundation, critical thinking, and problem-solving skills.

The ICSE Class 10 Math Chapter – Section and Midpoint formula is one of the most important concepts in Coordinate Geometry. They are used to find the coordinates of a point dividing a line segment or finding the midpoint of a line segment. Solving such problems allows students to understand the practical applications of the formulas and their significance in geometry.

Selina Solutions for Section and Midpoint Formula helps students prepare for their exams and future academic pursuits. They extensively cover the section and midpoint formulas and provide step-by-step explanations and solutions for each question in the textbook. Thus, students understand the underlying concepts and techniques required to apply the formulas. These solutions completely align with the ICSE exam pattern and marking scheme so that students can effectively prepare and improve their performance.

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Section and Mid-Point Formula Exercise Ex. 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 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 15

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 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 on the x-axis which divides AB in the ratio k : 1.

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

= 0

-6k + 3 = 0

k =


Hence, 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 .

(iii) Area of quadrilateral PMNQ

= (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)

 

 

Solution 26

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

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

Here m1 = 3 and m2 =1

 

Section and Mid-Point Formula Exercise Ex. 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 of the circle is the mid-point of diameter.

Let the required co-ordinates of the other end of diameter 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 are (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.

Section and Mid-Point Formula Exercise Ex. 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 AC 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 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 8

left parenthesis straight i right parenthesis space space space Radius space AC equals square root of open parentheses 3 plus 2 close parentheses squared plus open parentheses negative 7 minus 5 close parentheses squared end root
space space space space space space space space space space space space space space space space space space space space space space space space space space equals square root of 5 squared plus open parentheses negative 12 close parentheses squared end root
space space space space space space space space space space space space space space space space space space space space space space space space space space equals square root of 25 plus 144 end root
space space space space space space space space space space space space space space space space space space space space space space space space space space equals square root of 169
space space space space space space space space space space space space space space space space space space space space space space space space space space equals 13 space units
left parenthesis ii right parenthesis space space space Let space the space coordinates space of space straight B space be space left parenthesis straight x comma space straight y right parenthesis.
space space space space space space space Using space mid minus point space formula comma space we space have
space space space space space space space minus 2 equals fraction numerator 3 plus straight x over denominator 2 end fraction space space space space space and space space space 5 equals fraction numerator negative 7 plus straight y over denominator 2 end fraction
space space space space space space space rightwards double arrow negative 4 equals 3 plus straight x space space space and space space 10 equals negative 7 plus straight y
space space space space space space space rightwards double arrow straight x equals negative 7 space space and space space space straight y equals 17
space space space space space space space Thus comma space the space coordinates space of space straight B space are space left parenthesis negative 7 comma space 17 right parenthesis.

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:

1 equals fraction numerator 2 straight a minus 2 over denominator 2 end fraction space
rightwards double arrow 1 equals straight a minus 1 space
rightwards double arrow straight a equals 2
And space space 2 straight a plus 1 equals fraction numerator 4 plus 2 straight b over denominator 2 end fraction
rightwards double arrow 2 straight a plus 1 equals 2 plus straight b
rightwards double arrow 2 cross times 2 plus 1 equals 2 plus straight b
rightwards double arrow 5 equals 2 plus straight b
rightwards double arrow straight b equals 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

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 20

Solution 21

Solution 22

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

  

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