What are the co ordinates of the points of trisection of the line segment joining the points H 3 2 and H 3 4?

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Ex 7.2, 2 Find the coordinates of the points of trisection of the line segment joining (4, –1) and (–2, –3). Let the given points be A (4, −1) & B (−2, 3) P & Q are two points on AB such that AP = PQ = QB Let AP = PQ = QB = m Point P divides AP & PB in the ratio AP = m PB = PQ + QB = k + k = 2k Hence, Ratio between AP & PB = AP/PB = 𝑘/2𝑘 = 1/2 Thus P divides AB in the ratio 1:2 Finding P Let P(x, y) m1 = 1, m2 = 2 And for AB x1 = 4, x2 = −2 y1 = −1, y2 = −3 x = (𝑚_1 𝑥_2 + 𝑚_2 𝑥_1)/(𝑚_1 + 𝑚_2 ) = (1 ×−2 + 2 × 4)/(1 + 2) = (−2 + 8)/3 = 6/3 = 2 y = (𝑚_1 𝑦_(2 )+ 𝑚_2 𝑦_1)/(𝑚_1 + 𝑚_2 ) = (1 × −3 + 2 × −1)/(1 + 2) = (−3 − 2)/3 = (−5)/3 Hence, point P is P(x, y) = P ("2, " (−5)/3) Similarly, Point Q divides AB in the ratio AQ & QB 𝐴𝑄/𝑄𝐵 = (𝐴𝑃 + 𝑃𝑄)/𝑄𝐵 = (𝑘 + 𝑘 )/𝑘 = 2𝑘/𝑘 = 2/1 = 2 : 1 Finding Q Let Q be Q(x , y) m1 = 2, m2 = 1 x1 = 4, x2 = −2 y1 = −1, y2 = −3 x = (𝑚_1 𝑥_2 + 𝑚_2 𝑥_1)/(𝑚_1 + 𝑚_2 ) = (2 ×−2 + 1 × 4)/(2 + 1) = (−4 + 4)/3 = 0/3 = 0 y = (𝑚_1 𝑦_2 + 𝑚_2 𝑦_1)/(𝑚_1 + 𝑚_2 ) = (2 × −3 + 1 × −1)/(1 + 2) = (−6 − 1)/3 = (−7)/3 Hence, point Q is Q(x, y) = Q ("0, " (−𝟕)/𝟑)

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).

☛ Related Questions:

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Find the ratio in which the line segment joining the points (- 3, 10) and (6, - 8) is divided by (- 1, 6).
Find the ratio in which the line segment joining A (1, - 5) and B (- 4, 5) is divided by the x-axis. Also, find the coordinates of the point of division.
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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))`

Find the points of trisection of the line segment joining the points A 4,3 and B 2, 1.A. 2, 5/3 and 0, 1/3B. 2, 5/3 and 2, 1/3C. 2, 5/3 and 0, 1/3D. 2, 5/3 and 0, 4/3

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