Question

Let P and Q be the points of trisection of the line segment joining the points A( 2, -2 ) and B( - 7, 4 ) such that P is nearer to A. Find the coordinates of P and Q.

Answer 1

Answer:

⇒ P = (-1,0)

⇒ Q = (-4,2).

Step-by-step explanation:

Given points are A(2,-2) and B(-7,4).

Let P(x,y) and Q(a,b) be the points of trisection of AB so that AP = PQ = AB.

P divides AB in the ratio 1:2 and Q divides AB in the ratio 2:1.

Section formula:

⇒ lx₂ + mx₁)/l+m, (ly₂ + my₁)/l+m.

(i)

P divides the line segment in the ratio 1:2.

l = 1, m = 2.

A(2,-2) and B(-7,4)

⇒ [1(-7) +2(2)]/1 + 2, [1(4) + 2(-2)]/1 + 2.

⇒ [-7 + 4]/3, [0/3]

⇒ [-3/3, 0]

⇒ [-1,0]

(ii)

Q divides the line segment in the ratio 2:1

l = 1, m = 2.

A(2,-2) and B(-7,4)

⇒ [2(-7) + 1(2)]/3, [2(4) + 1(-2)]/1 + 2

⇒ [-14 + 2]/3, [8 - 2]/3

⇒ [-12/3], [6/3]

⇒ [-4,2]

Hope it helps!

Answer 2

According to the question : -

PA = PQ = QB { Due to trisection }

And, 2PA = PB

=> AP / PB = 1 / 2

By section formula
$Coordinates \:of \: P = \bigg( \dfrac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}} .\dfrac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}} \bigg) = \bigg(\dfrac{(1 \times - 7) + (2 \times 2)}{1 + 2},\dfrac{(1 \times 4) + (2 \times - 2)}{3} \bigg) \\ \\ \implies \bigg( - \dfrac{3}{3} ,\frac{0}{3} \bigg)$

Therefore, co ordinates of P = (- 1 , 0 )

Now PQ = QB
→ Q is the mid point of PB

by mid point formula :

$Co ordinates \: of \:Q = \bigg( \dfrac{x_{1}+x_{2}}{2} , \dfrac{y_{1}+y_{2}}{2} \bigg) = \bigg ( \dfrac{-1 - 7}{2} , \dfrac{0 + 4}{2} \bigg)$

Therefore, co ordinates of Q = ( - 4 , 2 ).