Question

the line segment joining the points (3,-4)and (1,2)is trisected at the point of p and q if the coordinates of p and q are (p,-2) and (5/3,q) find the value of p and q

Answer 1

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Answer 2

Answers:

p = 7/3

q = 0

Correction in the question:

The line segment joining the points (3, -4) and (1, 2) is trisected at the point of 'P' and 'Q'. If the coordinates of 'P' and 'Q' are (p ,-2) and (5/3, q) find the value of p and q .

Solution:

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

The points of trisection are P(p, -2) and Q(5/3, q).

$\setlength{\unitlength}{1cm}\begin{picture}(6,6)\put(1.85,3.9){\textbullet}\put(2, 4){\line(1, 0){6}}\put(1.4,3.5){\sf A(3, -4)}\put(2.7,4.5){\sf 1x}\put(3.3,3.5){\sf P(p, -2)}\put(4.8,4.5){\sf 1x}\put(3.83,3.9){\textbullet}\put(5.15,3.5){\sf Q(5/3, q)}\put(5.85,3.9){\textbullet}\put(6.8,4.5){\sf 1x}\put(7.4,3.5){\sf B(1, 2)}\put(7.9,3.9){\textbullet}\end{picture}$

Since they're trisected, the points P and Q divide the line AB into three equal parts AP, PQ and QB.

Therefore:

AP : PQ : QB = 1x : 1x : 1x

Considering the line AB with Q as the point of intersection.

$\setlength{\unitlength}{1cm}\begin{picture}(6,6)\put(1.85,3.9){\textbullet}\put(2, 4){\line(1, 0){6}}\put(1.4,3.5){\sf A(3, -4)}\put(3.8,4.5){\sf 2x}\put(5.15,3.5){\sf Q(5/3, q)}\put(5.85,3.9){\textbullet}\put(6.8,4.5){\sf 1x}\put(7.4,3.5){\sf B(1, 2)}\put(7.9,3.9){\textbullet}\end{picture}$

We know that:

AQ = AP + PQ

AQ = 1x + 1x

AQ = 2x

And we know that QB = 1x.

Therefore the line AB is divided in the ratio 2:1 by the point Q.

Using the section formula we get:

$\Longrightarrow \sf Q(x, y) = \Bigg[\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2} , \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2} \Bigg]$

Here:

m₁ = 2 and m₂ = 1

x₁ = 3 and y₁ = -4

x₂ = 1 and y₂ = 2

x = 5/3 and y = q

$\Longrightarrow \sf Q\Bigg[\dfrac{5}{3}, q\Bigg] = \Bigg[\dfrac{(2)(1) + (1)(3)}{2 + 1} , \dfrac{(2)(2) + (1)(-4)}{2 + 1} \Bigg]$

$\Longrightarrow \sf Q\Bigg[\dfrac{5}{3}, q\Bigg] = \Bigg[\dfrac{5}{3} , \dfrac{4 - 4}{3} \Bigg]$

$\Longrightarrow \sf Q\Bigg[\dfrac{5}{3}, q\Bigg] = \Bigg[\dfrac{5}{3} , 0\Bigg]$

Equating the y-coordinate of Q(5/3, q) with the y-coordinate of [5/3, 0] we get:

$\Longrightarrow \sf q = 0$

Therefore the value of 'q' is 0.

________________________

Considering the line AQ with P as the midpoint:

$\setlength{\unitlength}{1cm}\begin{picture}(6,6)\put(1.85,3.9){\textbullet}\put(2, 4){\line(1, 0){4}}\put(1.4,3.5){\sf A(3, -4)}\put(2.7,4.5){\sf 1x}\put(3.4,3.5){\sf P(p, -2)}\put(4.8,4.5){\sf 1x}\put(3.83,3.9){\textbullet}\put(5.18,3.5){\sf Q(5/3, 0)}\put(5.9,3.9){\textbullet}\end{picture}$

[P is the midpoint as it divides AQ in the ratio 1:1]

Using the midpoint formula we get:

$\Longrightarrow \sf P(x, y) = \Bigg[\dfrac{x_1 + x_2}{2} , \dfrac{y_1 + y_2}{2} \Bigg]$

Here:

x₁ = 3 and y₁ = -4

x₂ = 5/3 and y₂ = 0

x = p and y = -2

$\Longrightarrow \sf P(p, -2) = \Bigg[\dfrac{3 + \big[\frac{5}{3}\big]}{2} , \dfrac{-4 + 0}{2} \Bigg]$

$\Longrightarrow \sf P(p, -2) = \Bigg[\dfrac{\big[\frac{9 \ + \ 5}{3}\big]}{2} , \dfrac{-4}{2} \Bigg]$

$\Longrightarrow \sf P(p, -2) = \Bigg[\dfrac{\big[\frac{14}{3}\big]}{2} , -2 \Bigg]$

$\Longrightarrow \sf P(p, -2) = \Bigg[\dfrac{14}{6} , -2 \Bigg]$

$\Longrightarrow \sf P(p, -2) = \Bigg[\dfrac{7}{3} , -2 \Bigg]$

Equating the x-coordinate of P(p, -2) with the x-coordinate of [7/3, -2] we get:

$\Longrightarrow \sf p =\dfrac{7}{3}$

Therefore the value of 'p' is 7/3.

Hence solved.