CBSE Tutorials

Coordinate Geometry | Section Formula

Section formula :

The coordinates of the point P(x, y) which divides the line segment joining and internally in the ratio m : n are given by

Proof:    

Let X’OX and YOY’ be the coordinate axes.

Let and be the end points of the given line segment AB.

Let P(x, y) be the point which divides AB in the ratio m : n.

Then,    

Draw    AL ^ OX; BM ^ OX; PN ^ OX; AR ^ PN; and PS ^ BM

Now,    AR = LN = ON – OL =

PS = NM = OM – ON =

PR = PN – RN = PN – AL =

BS = MB – SM = BM – PN = .

DARP and DPSB are similar and, therefore, their sides are proportional.

\    

Þ    

Þ    

Þ     and

Þ     and

Þ    

Hence, the coordinates of P are .

Midpoint formula :

The coordinates of the midpoint M of a line segment AB with end points and are .

Proof:   

 Let M be the midpoint of the line segment joining the points and .

Then, M divides AB in the ratio 1 : 1

So, by the section formula, the coordinates of M are

, i.e., .

Hence, the coordinates of the midpoint of AB are .

Question:

Find the coordinates of the point which divides the line segment joining the points (6, 3) and (–4, 5) in the ratio 3 : 2 internally.

Solution:

Let P(x, y) be the required point. Then,

and
Þ x = 0 and y =

So, the coordinates of P are (0, 21/5).

Question:

In what ratio does the point C(3/5, 11/5) divide the line segment joining the points A(3, 5) and B(–3, –2)?.

Solution:

Let the point C divide AB in the ratio k : 1. Then the coordinates of C are

But, the coordinates of C are given as (3/5, 11/5).

\     and

Þ    – 15k + 15 = 3k + 3 and – 10k + 25 = 11k + 11

Þ    18k = 12 and 21k = 14

Þ    . Hence, the point C divides AB in the ratio 2 : 3.

Question:

Find the coordinates of the points of trisection (i.e., points dividing in three equal parts) of the line segment joining the points A(2, –2) and B(–7, 4).

Solution:

Let P and Q be the points of trisection of AB i.e., AP = PQ = QB.

Therefore, P divides AB internally in the ratio 1 : 2. Therefore, the coordinates of P, by applying the section formula, are

, i.e., (–1, 0)

Now, Q also divides AB internally in the ratio 2 : 1. So, the coordinates of Q are

, i.e., (–4, 2)

Therefore, the coordinates of the points of trisection of the line segment joining A and B are (–1, 0) and (–4, 2).

Question:

Find the ratio in which the y-axis divides the line segment joining the points (5, –6) and (–1, –4). Also find the point of intersection.

Solution:

Let the ratio be k : 1. Then by the section formula, the coordinates of the point which divides AB in the ratio k : 1 are .

This point lies on the y-axis, \ its coordinates are (0, y)

\         

So,        k = 5

Thus, the ratio is 5 : 1.

Putting the value of k = 5, we get the point of intersection as .

Question:

In what ratio does the x-axis divide the line segment joining the points (2, –3) and (5, 6)? Also, find the coordinates of the point of intersection.

Solution:

Let the required ratio be k : 1. Then, the coordinates of the point R which divides PQ in the ratio k : 1 are

This point lies on x-axis therefore its coordinates are (x, 0)

\    . Thus, the required ratio is or 1 : 2.

Putting k = 1/2 in the coordinates of R, we find that its coordinates are (3, 0).

Question:

If the points A(6, 1), B(8, 2), C(9, 4) and D(p, 3) are the vertices of a parallelogram, taken in order, find the value of p.

Solution:

In parallelogram ABCD

coordinates of the mid-point of AC = coordinates of the mid-point of BD

Þ    

Þ    

Þ    

Þ    30 = 16 + 2p

Þ    14 = 2p

Þ     p = 7

Question:

The three vertices of a parallelogram taken in order are (–1, 0), (3, 1) and (2, 2) respectively. Find the coordinates of the fourth vertex.

Solution:

Let A(–1, 0), B(3, 1), C(2, 2) and D(x, y) be the vertices of a parallelogram ABCD taken in order. Since, the diagonals of a parallelogram bisect each other.

\    Coordinates of the mid-point of AC = Coordinates of the mid-point of BD

Þ    

Þ     

Þ     and

Þ     x = – 2 and y = 1

Hence, the fourth vertex of the parallelogram is (–2, 1).

Question:

Find the ratio in which the point (–3, p) divides the line segment joining the points (–5, –4) and (–2, 3). Hence, find the value of p.

Solution:

Suppose the point P(–3, p) divides the line segment joining points A(–5, –4) and B(–2, 3) in the ratio k : 1.

Then, the coordinates of P are

But, the coordinates of P are given as (–3, p).

\     and

Þ    –2k – 5 = –3k – 3 and

Þ    k = 2 and p =

Þ    k = 2 and p = .

Hence, the ratio is 2 : 1 and p = .

Question:

The coordinates of one end point of a diameter AB of a circle are A(4, –1) and the coordinates of the centre of the circle are C(1, –3). Find the coordinates of B.

Solution:

In the figure A(4, –1) and B(a, b) are the end points of the given diameter AB, and C(1, –3) is the centre of the circle. Then, C is the midpoint of AB.

By the midpoint formula, the coordinates of C are .

But, the coordinates of C are given as (1, –3).

\    

Þ    4 + a = 2 and –1 + b = –6

Þ      a = –2 and b = –5

Hence, the required point is B(–2, –5).

Question:

If A(5, –1), B(–3, –2) and C(–1, 8) are the vertices of triangle ABC, find the length of median through A.

Solution:

Let AD be the median through the vertex A of DABC. Then, D is the midpoint of BC.

By midpoint formula, the coordinates of D are i.e., (–2, 3).

\    

units

Question:

If the coordinates of the midpoints of the sides of a triangle are (1, 2)(0, –1) and (2, –1). Find the coordinates of its vertices.

Solution:

Let A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) be the vertices of DABC. Let D (1, 2), E(0, –1), and F(2, –1) be the midpoints of sides BC, CA and AB respectively. Since D is the midpoint of BC.

\     and

Þ    x₂ + x₃ = 2 and y₂ + y₃ = 4        …(i)

Similarly, E and F are midpoints of CA and AB respectively.

\     and

Þ    x₁ + x₃ = 0 and y₁ + y₃ = – 2        …(ii)

and,     and

Þ    x₁ + x₂ = 4 and y₁ + y₂ = – 2        …(iii)

From (i), (ii) and (iii), we get

(x₂ + x₃) + (x₁ + x₃) + (x₁ + x₂) = 2 + 0 + 4 and

(y₂ + y₃) + (y₁ + y₃) + (y₁ + y₂) = 4 – 2 – 2

Þ    2(x₁ + x₂ + x₃) = 6 and 2(y₁ + y₂ + y₃) = 0

Þ    x₁ + x₂ + x₃ = 3 and y₁ + y₂ + y₃ = 0    …(iv)

Þ    From (i) and (iv), we get

x₁ + 2 = 3 and y₁ + 4 = 0

Þ    x₁ = 1 and y₁ = – 4. So, the coordinates of A are (1, – 4)

From (ii) and (iv), we get

x₂ + 0 = 3 and y₂ – 2 = 0

Þ     x₂ = 3 and y₂ = 2. So, coordinates of B are (3, 2).

From (iii) and (iv), we get

x₃ + 4 = 3 and y₃ – 2 = 0

Þ     x₃ = – 1 and y₃ = 2. So, coordinates of C are (–1, 2).

Hence, the vertices of the triangle ABC are A(1, – 4), B(3, 2) and C(–1, 2).

Centroid of a Triangle

The coordinates of the centroid of a triangle with vertices and is given by .

Proof:

Let and

be the vertices of a DABC.

Let D be the midpoint of BC

Then, the coordinates of D by midpoint formula are .

Let G(x, y) be the centroid of DABC.

Then, G divides AD in the ratio 2 : 1

\    

Hence, the coordinates of G are .

Question:

Find the centroid of DABC whose vertices are A(–3, 0), B(5, –2) and C(–8, 5).

Solution:

Here, and

Let G(x, y) be the centroid of DABC. Then,

Hence, the centroid of DABC is G(–2, 1)

Question:

Two vertices of a DABC are given by A(6, 4) and B(–2, 2), and its centroid is G(3, 4). Find the coordinates of the third vertex C of DABC.

Solution:

Two vertices of DABC are A(6, 4) and B(–2, 2). Let the third vertex by C(a, b).

Then, the coordinates of its centroid are

, i.e.,

But, it is given that the centroid is G(3, 4).

\     and

Þ    a = 5 and b = 6

Hence, the third vertex of DABC is C(5, 6)