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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 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

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 13

Solution 14

Solution 15

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 17

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 .

 

(iii) Area of quadrilateral PMNQ

= (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.

Solution 21

(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:

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