# CBSE 10th Mathematics | Coordinate Geometry | Section Formula

## Coordinate Geometry | Section Formula

### Section formula :

The coordinates of the point P(x, y) which divides the line segment joining $\displaystyle A({{x}_{1}},\ {{y}_{1}})$ and $\displaystyle B({{x}_{2}},\ {{y}_{2}})$ internally in the ratio m : n are given by

$\displaystyle x=\frac{m{{x}_{2}}+n{{x}_{1}}}{m+n},\ y=\frac{m{{y}_{2}}+n{{y}_{1}}}{m+n}$

Proof:

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

Let $\displaystyle A({{x}_{1}},\ {{y}_{1}})$ and $\displaystyle B({{x}_{2}},\ {{y}_{2}})$ 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,    $\displaystyle \frac{AP}{PB}=\frac{m}{n}$

Draw    AL OX; BM OX; PN OXAR PN; and PS BM

Now,    AR = LN = ONOL = $\displaystyle (x-{{x}_{1}});$

PS = NM = OMON = $\displaystyle ({{x}_{2}}-x);$

PR = PNRN = PNAL = $\displaystyle (y-{{y}_{1}});$

BS = MBSM = BMPN = $\displaystyle ({{y}_{2}}-y)$.

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

\    $\displaystyle \frac{AP}{PB}=\frac{AR}{PS}=\frac{PR}{BS}$

Þ    $\displaystyle \frac{m}{n}=\frac{x-{{x}_{1}}}{{{x}_{2}}-x}=\frac{y-{{y}_{1}}}{{{y}_{2}}-y}$

Þ    $\displaystyle \frac{m}{n}=\frac{x-{{x}_{1}}}{{{x}_{2}}-x}\ \ \text{and}\ \ \frac{m}{n}=\frac{y-{{y}_{1}}}{{{y}_{2}}-y}$

Þ    $\displaystyle m{{x}_{2}}-mx=nx-n{{x}_{1}}$ and $\displaystyle m{{y}_{2}}-my=ny-n{{y}_{1}}$

Þ    $\displaystyle (m+n)x=(m{{x}_{2}}+n{{x}_{1}})$ and $\displaystyle (m+n)y=(m{{y}_{2}}+n{{y}_{1}})$

Þ    $\displaystyle x=\frac{m{{x}_{2}}+n{{x}_{1}}}{m+n},\ y=\frac{m{{y}_{2}}+n{{y}_{1}}}{m+n}$

Hence, the coordinates of P are $\displaystyle \left( \frac{m{{x}_{2}}+n{{x}_{1}}}{m+n},\ \frac{m{{y}_{2}}+n{{y}_{1}}}{m+n} \right)$.

### Midpoint formula :

The coordinates of the midpoint M of a line segment AB with end points $\displaystyle A({{x}_{1}},\ {{y}_{1}})$ and $\displaystyle B({{x}_{2}},\ {{y}_{2}})$ are $\displaystyle M\,\left( \frac{{{x}_{1}}+{{x}_{2}}}{2},\ \frac{{{y}_{1}}+{{y}_{2}}}{2} \right)$.

Proof:

Let M be the midpoint of the line segment joining the points $\displaystyle A\,({{x}_{1}},\ {{y}_{1}})$ and $\displaystyle B\,({{x}_{2}},\ {{y}_{2}})$.

Then, M divides AB in the ratio 1 : 1

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

$\displaystyle \left( \frac{1\cdot {{x}_{2}}+1\cdot {{x}_{1}}}{1+1},\ \frac{1\cdot {{y}_{2}}+1\cdot {{y}_{1}}}{1+1} \right)$, i.e., $\displaystyle \left( \frac{{{x}_{1}}+{{x}_{2}}}{2},\ \frac{{{y}_{1}}+{{y}_{2}}}{2} \right)$.

Hence, the coordinates of the midpoint of AB are $\displaystyle \left( \frac{{{x}_{1}}+{{x}_{2}}}{2},\ \frac{{{y}_{1}}+{{y}_{2}}}{2} \right)$.

### Solution:

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

$\displaystyle x=\frac{3\times -4+2\times 6}{3+2}$ and $\displaystyle y=\frac{3\times 5+2\times 3}{3+2}$
Þ x = 0 and y = $\displaystyle \frac{21}{5}$

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

### Solution:

Let the point C divide AB in the ratio k : 1. Then the coordinates of C are $\displaystyle \left( \frac{-3k+3}{k+1},\ \frac{-2k+5}{k+1} \right)$

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

\    $\displaystyle \frac{-3k+3}{k+1}=\frac{3}{5}$ and $\displaystyle \frac{-2k+5}{k+1}=\frac{11}{5}$

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

Þ    18k = 12 and 21k = 14

Þ    $\displaystyle k=\frac{2}{3}$. Hence, the point C divides AB in the ratio 2 : 3.

### 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

$\displaystyle \left( \frac{1(-7)+2(2)}{1+2},\ \frac{1(4)+2(-2)}{1+2} \right)$, i.e., (–1, 0)

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

$\displaystyle \left( \frac{2(-7)+1(2)}{2+1},\ \frac{2(4)+1(-2)}{2+1} \right)$, 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).

### 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 $\displaystyle \left( \frac{-k+5}{k+1},\ \frac{-4k-6}{k+1} \right)$.

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

\         $\displaystyle \frac{-k+5}{k+1}=0$

So,        k = 5

Thus, the ratio is 5 : 1.

Putting the value of k = 5, we get the point of intersection as $\displaystyle \left( 0,\ \frac{-13}{3} \right)$.

### Solution:

Let the required ratio be k : 1. Then, the coordinates of the point R which divides PQ in the ratio k : 1 are $\displaystyle \,\left( \frac{5k+2}{k+1},\ \frac{6k-3}{k+1} \right)$

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

\    $\displaystyle \frac{6k-3}{k+1}=0\Rightarrow \frac{1}{2}$. Thus, the required ratio is $\displaystyle \frac{1}{2}\ :\ 1$ or 1 : 2.

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

### Solution:

In parallelogram ABCD

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

Þ    $\displaystyle \left( \frac{6+9}{2},\ \frac{1+4}{2} \right)=\left( \frac{8+p}{2},\ \frac{2+3}{2} \right)$

Þ    $\displaystyle \left( \frac{15}{2},\ \frac{5}{2} \right)=\left( \frac{8+p}{2},\ \frac{5}{2} \right)$

Þ    $\displaystyle \frac{15}{2}=\frac{8+p}{2}$

Þ    30 = 16 + 2p

Þ    14 = 2p

Þ     p = 7

### 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    $\displaystyle \left( \frac{-1+2}{2},\ \frac{0+2}{2} \right)=\left( \frac{3+x}{2},\ \frac{1+y}{2} \right)$

Þ     $\displaystyle \left( \frac{1}{2},\ 1 \right)=\left( \frac{3+x}{2},\ \frac{y+1}{2} \right)$

Þ    $\displaystyle \frac{3+x}{2}=\frac{1}{2}$ and $\displaystyle \frac{y+1}{2}=1$

Þ     x = – 2 and y = 1

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

### 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 $\displaystyle \left( \frac{-2k-5}{k+1},\ \frac{3k-4}{k+1} \right)$

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

\    $\displaystyle \frac{-2k-5}{k+1}=-3$ and $\displaystyle \frac{3k-4}{k+1}=p$

Þ    –2k – 5 = –3k – 3 and $\displaystyle \frac{3k-4}{k+1}=p$

Þ    k = 2 and p = $\displaystyle \frac{3k-4}{k+1}$

Þ    k = 2 and p = $\displaystyle \frac{2}{3}$.

Hence, the ratio is 2 : 1 and p = $\displaystyle \frac{2}{3}$.

### 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 $\displaystyle \left( \frac{4+a}{2},\ \frac{-1+b}{2} \right)$.

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

\    $\displaystyle \frac{4+a}{2}=1\ \text{and}\ \frac{-1+b}{2}=-3$

Þ    4 + a = 2 and –1 + b = –6

Þ      a = –2 and b = –5

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

### 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 $\displaystyle \left( \frac{-3-1}{2},\ \frac{-2+8}{2} \right)$ i.e., (–2, 3).

\    $\displaystyle AD=\sqrt{{{(5+2)}^{2}}+{{(-1-3)}^{2}}}$

$\displaystyle =\sqrt{49+16}=\sqrt{65}$units

### 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.

\    $\displaystyle \frac{{{x}_{2}}+{{x}_{3}}}{2}=1$ and $\displaystyle \frac{{{y}_{2}}+{{y}_{3}}}{2}=2$

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

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

\    $\displaystyle \frac{{{x}_{1}}+{{x}_{3}}}{2}=0$ and $\displaystyle \frac{{{y}_{1}}+{{y}_{3}}}{2}=-1$

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

and,    $\displaystyle \frac{{{x}_{1}}+{{x}_{2}}}{2}=2$ and $\displaystyle \frac{{{y}_{1}}+{{y}_{2}}}{2}=-1$

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 $\displaystyle ({{x}_{1}},{{y}_{1}}),\,\,({{x}_{2}},{{y}_{2}})$ and $\displaystyle ({{x}_{3}},{{y}_{3}})$ is given by $\displaystyle \frac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}}{3},\ \frac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}}{3}$.

Proof:

Let $\displaystyle A({{x}_{1}},\ {{y}_{1}}),\ B({{x}_{2}},\ {{y}_{2}})$ and

$\displaystyle C({{x}_{3}},\ {{y}_{3}})$ be the vertices of a DABC.

Let D be the midpoint of BC

Then, the coordinates of D by midpoint formula are $\displaystyle \ \left( \frac{{{x}_{2}}+{{x}_{3}}}{2},\ \frac{{{y}_{2}}+{{y}_{3}}}{2} \right)$.

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

Then, G divides AD in the ratio 2 : 1

\    $\displaystyle x=\left\{ \frac{2\cdot \frac{({{x}_{2}}+{{x}_{3}})}{2}+1\cdot {{x}_{1}}}{2+1} \right\}=\frac{({{x}_{1}}+{{x}_{2}}+{{x}_{3}})}{3}$

$\displaystyle y=\left\{ \frac{2\cdot \frac{({{y}_{2}}+{{y}_{3}})}{2}+1\cdot {{y}_{1}}}{2+1} \right\}=\frac{({{y}_{1}}+{{y}_{2}}+{{y}_{3}})}{3}$

Hence, the coordinates of G are $\displaystyle \left( \frac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}}{3},\ \frac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}}{3} \right)$.

### Solution:

Here, $\displaystyle {{x}_{1}}=-3,\ {{y}_{1}}=0,\ {{x}_{2}}=5,\ {{y}_{2}}=-2$ and $\displaystyle {{x}_{3}}=-8,\ {{y}_{3}}=5$

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

$\displaystyle x=\frac{1}{3}({{x}_{1}}+{{x}_{2}}+{{x}_{3}})=\frac{1}{3}(-3+5-8)=-2$

$\displaystyle y=\frac{1}{3}({{y}_{1}}+{{y}_{2}}+{{y}_{3}})=\frac{1}{3}(0-2+5)=1$

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

### 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

$\displaystyle G\,\left( \frac{6-2+a}{3},\ \frac{4+2+b}{3} \right)$, i.e., $\displaystyle G\,\left( \frac{4+a}{3},\ \frac{6+b}{3} \right)$

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

\    $\displaystyle \frac{4+a}{3}=3$ and $\displaystyle \frac{6+b}{3}=4$

Þ    a = 5 and b = 6

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

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