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).

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,

\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).

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 \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.

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

\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).

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 \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).

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 \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).

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    \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

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    \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).

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 \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}.

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 \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).

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 \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

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.

\    \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).

Question:

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

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)

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

\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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