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

(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.

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.

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.

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

Using section formula, we have:

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:

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 .

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:

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:

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

5AP = 2BP

The co-ordinates of the point P are

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).

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.

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).

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

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.

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

(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.

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.

(i) Co-ordinates of P are

Co-ordinates of Q are

(ii) Using distance formula, we have:

BC =

PQ =

Hence, PQ = BC.

BP: PC = 2: 3

Co-ordinates of P are

Using distance formula, we have:

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 .

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 .

(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

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)

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

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

Here m₁ = 3 and m₂ =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 =

Mid-point of AB = (2, 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:

(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).

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.

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.

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,

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).

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).

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.

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).

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).

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).

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

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.

Co-ordinates of the centroid of triangle ABC are

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)

Given, centroid of triangle ABC is the origin.

Given, BP: PC = 3: 2

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

Using distance formula, we have:

Using section formula,

Given, AB = 6AQ

Using section formula,

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.

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).

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.

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).

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).

Co- ordinates of the centroid of triangle ABC are

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

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.

(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.

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).

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.

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).  

P, Q and R are the mid points of the sides BC, CA and AB.

By mid - point formula, we get

From (i) and (ii), we get

Centroid of a ∆ABC = Centroid of a ∆PQR