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In the adjoining figure, P(5, -3) and Q(3, y) are the points of trisection of the line segment joining A(7, -2) and B(1, -5). Then, y equals :

In the adjoining figure, P(5, -3) and Q(3, y) are the points of trisection of the line segment joining A(7, -2) and B(1, -5). Then, y equals : Reflection, RSA Mathematics Solutions ICSE Class 10.

  1. -4

  2. (−52)\Big(\dfrac{-5}{2}\Big)

  3. 2

  4. 4

Section Formula

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Answer

Since P and Q trisect the line segment AB, the point Q(3, y) divides A(7, -2) and B(1, -5) in the ratio 2 : 1.

Let point Q be (3, y).

Given,

m1 : m2 = 2 : 1

By section-formula,

(x, y) = (m1x2+m2x1m1+m2,m1y2+m2y1m1+m2)\Big(\dfrac{m1x2 + m2x1}{m1 + m2}, \dfrac{m1y2 + m2y1}{m1 + m2}\Big)

Substituting values we get :

⇒(3,y)=(2×1+1×72+1,2×(−5)+1×(−2)2+1)⇒(3,y)=(2+73,−10−23)⇒(3,y)=(93,−123)⇒(3,y)=(3,−4).\Rightarrow (3, y) = \Big(\dfrac{2 \times 1 + 1 \times 7}{2 + 1}, \dfrac{2 \times (-5) + 1 \times (-2)}{2 + 1}\Big) \\[1em] \Rightarrow (3, y) = \Big(\dfrac{2 + 7}{3}, \dfrac{-10 - 2}{3}\Big) \\[1em] \Rightarrow (3, y) = \Big(\dfrac{9}{3}, \dfrac{-12}{3}\Big) \\[1em] \Rightarrow (3, y) = (3, -4).

Thus, y = -4.

Hence, Option 1 is the correct option.

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