What are the co ordinates of the points of trisection of the line segment joining the points 3 2 & (- 3 4?

Solution:

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 is given by the section formula.

Let the points be A(4, - 1) and B(- 2, - 3).

Let P (x₁, y₁) and Q (x₂, y₂) be the points of trisection of the line segment joining the given points.

Then, AP = PC = CB

By Section formula ,

P(x, y) = [(mx₂ + nx₁) / (m + n), (my₂ + ny₁) / (m + n)] ...... (1)

Considering A(4, - 1) and B(- 2, - 3), by observation point P(x₁, y₁) divides AB internally in the ratio 1 : 2.

Hence m : n = 1 : 2

By substituting the values in the Equation (1)

x₁ = [1 × (- 2) + 2 × 4] / (1 + 2)

x₁ = (- 2 + 8) / 3

= 2

y₁ = [1 × (- 3) + 2 × (- 1)] / (1 + 2)

y₁ = (- 3 - 2) / (1 + 2)

= - 5/3

Hence, P(x₁ , y₁) = (2, - 5/3)

Now considering A(4, - 1) and B(- 2, - 3), by observation point C(x₂, y₂) divides AB internally in the ratio 2 : 1.

Hence m : n = 2 : 1

By substituting the values in the Equation (1)

x₂ = [2 × (- 2) + 1 × 4] / (2 + 1)

= (- 4 + 4) / 3

= 0

y₂ = [2 × (- 3) + 1 × (- 1)] / (2 + 1)

= (- 6 - 1) / 3

= - 7/3

Therefore, C(x₂ , y₂) = (0, - 7/3)

Hence, the points of trisection are P(x₁ , y₁) = (2, - 5/3) and C (x₂ , y₂) = (0, - 7/3)

☛ Check: NCERT Solutions Class 10 Maths Chapter 7

Video Solution:

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

NCERT Class 10 Maths Solutions Chapter 7 Exercise 7.2 Question 2

Summary:

The coordinates of the points of trisection of the line segment joining (4, - 1) and (- 2, - 3) are (2, - 5/3) and (0, - 7/3).

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

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Text Solution

Solution : Let P `(a, b) and Q(c, d)` trisect the line joining the points `A(3, -2) and B(-3, -4)`. Now, point P (a, b) divides the line AB in the ratio `1: 2`. <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/NTN_MATH_X_C07_S01_035_S01.png" width="80%"> `therefore" "a=(1(-3)+2(3))/(1+2)=(3)/(3)=1` `" "b=(1(-4)+2(-2))/(1+2)=-(8)/(3)` Therefore, co-ordinates of point P = `(1, -(8)/(3))` `Q(c, d)` divides the line AB in the ratio 2 :1. `therefore" "c=(2(-3)+1(3))/(2+1)=-1` `" "d=(2(-4)+1(-2))/(2+1)=(-10)/(3)` Therfore, co-ordinates of `Q=(-1, (-10)/(3))` `therefore` Co-ordinates of points of trisection of AB = `=(1, (-8)/(3))and (-1, (-10)/(3))`

Let A(3, –2) and B(–3, –4) be the two given points.

Suppose P(x1, y1) and Q(x2, y2) are the points of trisection of the line segment joining the given points i.e. AP = PQ = QB.

Now,

PB = PQ + QB = AP + AP = 2AP

∴ AP : PB = AP : 2AP = 1 : 2

So, point P divides AB internally in the ratio 1 : 2.

Similarly,

AQ : QB = 2 : 1

P divides AB internally in the ratio 1 : 2.

\[\therefore \left( \frac{1 \times \left( - 3 \right) + 2 \times 3}{1 + 2}, \frac{1 \times \left( - 4 \right) + 2 \times \left( - 2 \right)}{1 + 2} \right) = \left( x_1 , y_1 \right)\]

\[ \Rightarrow \left( \frac{- 3 + 6}{3}, \frac{- 4 - 4}{3} \right) = \left( x_1 , y_1 \right)\]

\[ \Rightarrow \left( 1, - \frac{8}{3} \right) = \left( x_1 , y_1 \right)\]

\[ \Rightarrow x_1 = 1, y_1 = - \frac{8}{3}\]

Q divides AB internally in the ratio 2 : 1.

\[\therefore \left( \frac{2 \times \left( - 3 \right) + 1 \times 3}{1 + 2}, \frac{2 \times \left( - 4 \right) + 1 \times \left( - 2 \right)}{1 + 2} \right) = \left( x_2 , y_2 \right)\]

\[ \Rightarrow \left( \frac{- 6 + 3}{3}, \frac{- 8 - 2}{3} \right) = \left( x_2 , y_2 \right)\]

\[ \Rightarrow \left( - 1, - \frac{10}{3} \right) = \left( x_2 , y_2 \right)\]

\[ \Rightarrow x_2 = - 1, y_2 = - \frac{10}{3}\]

Thus, the coordinates of the points of trisection of the line segment joining the given points are

\[\left( 1, - \frac{8}{3} \right) and \left( - 1, - \frac{10}{3} \right) .\]