Coordinate Geometry

Solution of NCERT Exercise 7.2

Section Formula

The coordinates of the point P(x, y) which divides the line segment joining the points A(x₁, y₁) and B(x₂, y₂), internally, in the ratio m₁:m₂ are

`((m_1x_2+m_2x_1)/(m_1+m_2), (m_1y_2+m_2y_1)/(m_1+m_2))`

This is known as Section Formula

Special Case of Section Formula

The mid-point of a line segment divides the line segment in the ratio 1:1. Therefore, the coordinates of the mid-point P of the join of the points A(x₁, y₁) and B(x₂, y₂) is

`((1.x_1+1.x_2)/(1+1), (1.y_1+1.y_2)/(1+1))` `=((x_1x_2)/2, (y_1+y_2)/2)`

Question (1) Find the coordinates of the point which divides the join of (–1, 7) and (4, –3) in the ratio 2:3.

Solution

Given, points are (–1, 7) and (4, –3)

Ratio in which given divides the given join = 2:3

Thus, coordinates of point which divides the join = ?

Let the point P(x, y) divides the line joining the given points.

We know that,

According to section formula, if the coordinates of the point P(x, y) divides the line segment joining the points A(x₁, y₁) and B(x₂, y₂), internally, in the ratio m₁:m₂ , then

`x=(m_1x_2+m_2x_1)/(m_1+m_2)` and `y=(m_1y_2+m_2y_1)/(m_1+m_2)`

Here, we have, x₁ = –1, y₁= 7

And, x₂ = 4, y₂ =–3

And, m₁ = 2, m₂ =3

Thus, Using Section Formula, we have

`x=((2xx4)+(3xx(-1)))/(2+3)`

`=>x=(8-3)/5`

`=>x = 5/5=1`

And, `y=((2xx(-3))+(3xx7))/(2+3)`

`=>y=(-6+21)/5`

`=>y=15/5=3`

Thus, coordinates of point which divides the given line is (1, 3) Answer

Question (2) Find the coordinates of the points of trisection of the line segment joining (4, –1) and (–2, –3).

Solution:

Given points are (4, –1) and (–2, –3)

Thus, to find the coordinates of points of trisection of the line joining given two points.

Let, the given points are A(4, –1) and B(–2, –3)

And two points which divides the line in three parts are

P(x₁, y₁) and Q(x₂, y₂)

And, here AP = PQ = QB

We know that,

According to section formula, if the coordinates of the point P(x, y) divides the line segment joining the points A(x₁, y₁) and B(x₂, y₂), internally, in the ratio m₁:m₂ , then

`x=(m_1x_2+m_2x_1)/(m_1+m_2)` and `y=(m_1y_2+m_2y_1)/(m_1+m_2)`

(a) Coordinate for point P(x₁, y₁) which divides the line joining the line AB in 1:2 ratio

Here we have, A(x₁=4, y₁=–1)

And, B(x₂=–2, y₂=–3)

And, m₁=1 and m₂=2

According to section formula, For, P(x₁, y₁)

`x_1 = ((1xx(-2))+(2xx4))/(1+2)`

`=>x_1=(-2+8)/3`

`=>x_1=6/3`

`=>x_1=2`

And, `y_1=((1xx(-3))+(2xx(-1)))/(1+2)`

`=>y_1=(-3+(-2))/3`

`=>y_1=(-3-2)/3`

`=>y_1=(-5)/3`

Thus, P(x₁, y₁) `=(2, (-5)/3)`

(b) Coordinate for point Q(x₂, y₂) which divides the line joining the line AB in 2:1 ratio

Here we have, A(x₁=4, y₁=–1)

And, B(x₂=–2, y₂=–3)

And, m₁=2 and m₂=1

According to section formula, For Q(x₂, y₂

`x_2=((2x(-2))+(1xx4))/(2+1)`

`=>x_2=(-4+4)/3`

`=>x_2=0`

And, `y_2=((2xx(-3))+(1xx(-1)))/(2+1)`

`=>y_2=(-6+(-1))/3`

`=>y_2=(-7)/3`

Thus, coordinates of required points are `=(2, (-5)/3)` and `(0, (-7)/3)` Answer

Question (3) To conduct Sports Day activities, in your rectangular shaped school ground ABCD, line have been drawn with chalk powder at a distance of 1m each. 100 flower pots have been placed at a distance of 1m from each other along AD, as shown in figure. Niharika runs `1/4\(th)` of the distance AD on the 2nd line and posts a green flag. Preet runs `1/5\(th)` the distance AD on the eighth line and posts a red flag. What is the distance between both the flats? If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her flag?

Solution

Total number of flower pots = 100

Number of flower pots along the line AD = 100

Distance between each flower pots = 1 m

Thus, total distance along one line = 100 × 1m = 100m

Niharika post the flag at `1/4\(th)` of the distance AD on the 2nd line.

Preet runs `1/5\(th)` the distance AD on the eighth line

Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags.

Then, find the distance between two flags and coordinate of point exactly halfway = ?

Let, point at which Niharika posts the flag is `N` and point at which Preet posts the flag is `P`

And, point exactly halfway between N and P is H

Coordinate for point N

Since Niharik posts a flag at `1/4` th of the distance of AD on the second line

Thus, distance `=1/4xx100\ m` = 25 m

And, line number = 2

Thus, coordinate of point N = (2, 25)

Coordinate of Point P

Since Niharik posts a flag at `1/5` th of the distance of AD on the eighth line

Thus, distance `=1/5xx100\ m` = 20 m

And, line number = 8

Thus, coordinate of point P = (8, 20)

Finding the distance between N(2, 25) and P(8, 20)

Now, we know that, in coordinate geometry according to distance formula, in a plane the distance between two given points `=sqrt((x_2-x_1)^2+(y_2-y_1)^2)`

Here, we have, x₁ = 2, y₁ = 25

And, x₂ = 8, and y₂ = 20

Thus, using distance formula,

NP `=sqrt((8-2)^2+(20-25)^2)`

`=sqrt(6^2+(-5)^2)`

`=sqrt(36+25)`

⇒ NP `=sqrt(61)` m

Finding the coordinate of point R where flag post by Rashmi exactly between two flags

Let the coordinate of R = x, y

And we have, x₁ = 2, y₁ = 25

And, x₂ = 8, and y₂ = 20

Thus, `x=(x_1+x_2)/2`

`= (2+8)/2 =10/2`

`=>x=5`

And `y=(y_1+y_2)/2`

`=(25+20)/2`

`=45/2`

⇒ `y=45/2`

Thus, coordinate of mid-point between two flags is `(5, 45/2)`

Thus, distance between two flags `=sqrt(61)` and coordinate of mid-point between two flags `=(5, 45/2)` Answer

Question (4) Find the ratio in which the line segment joining the points (–3, 10) and (6, –8) is divided by (–1, 6).

Solution

Given, points are (–3, 10) and (6, –8)

And points that divide the given line in a ratio is (–1, 6)

Thus, ratio in which the given point divide the line = ?

We know that, If the ratio in which a point P divides the line AB is k:1, then the coordinates of the point P will be `((kx_2+x_1)/(k+1), (ky_2+y_1)/(k+1))`

Here we have, x₁ = –3, x₂ =6

And, y₁ = 10 and y₂ = –8

And, coordinate of point which divides the line is P(–1, 6)

Thus, x=–1 and y = 6

Let the point divides the line in k:1 ratio

Thus,

`x=-1=((kxx6+(-3))/(k+1))`

`=>-1=((6k-3)/(k+1))`

After cross multiplication

⇒ –1(k + 1) = 6k – 3

⇒ –k –1 = 6k – 3

⇒ –k + 6 k = –3 + 1

⇒ –7 k = – 2

`=>k = 2/7`

And, `y=6=((kxx(-8)+10)/(k+1))`

`=> 6 = ((-8k+10)/(k+1))`

After cross multiplication

⇒ 6 (k + 1) = –8k + 10

⇒ 6 k + 6 = –8k + 10

⇒ 6k + 8k = 10 – 6

⇒ 14 k = 4

`=>k = 4/14`

`=>k = 2/7`

Since, ratio is k : 1

Thus, after substituting the value of k, we have the required ratio

`2/7:1`

`=>2/7=1`

Or, 2 : 7

Thus, required ratio is 2 : 7 Answer

Question (5) Find the ratio in which the line segment joining A(1, –5) and B(–4, 5) is divided by x–axis. Also find the coordinates of the point of division.

Solution

Given, joining point of line segment are A(1, –5) and B(–4, 5)

Thus, ratio in which line divided by x–axis and coordinate of point which divides the line = ?

Let the ratio in which x–axis divides the line is k:1

We know that, If the ratio in which a point P divides the line AB is k:1, then the coordinates of the point P will be `((kx_2+x_1)/(k+1), (ky_2+y_1)/(k+1))`

Here, we have, x₁ = 1, x₂ = –4

And, y₁ = –5, y₂ = 5

Let the coordinate of point which divides the line is (x, y)

Since line x–axis divide the, thus, y = 0

Thus, `y=((kxx5+(-5))/(k+1))`

`=>0=((5k-5)/(k+1))`

After cross multiplication

⇒ 0 (k+1) = 5k – 5

⇒ 5k – 5 = 0

⇒ 5k = 5

Thus, `k=5/5=1`

Thus, required ratio is 1:1

Now, `x=((kxx(-4))+1)/(1+1)`

`=>x=((1xx(-4)+1)/2)`

`=>x = ((-4+1)/2)`

`=>x =(-3)/2`

Thus, ratio is 1:1 and coordinate of dividing point `=((-3)/2\, 0)` Answer

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