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The point P(-4, 1) divides the line segment joining the points A(2, -2) and B in the ratio 3 : 5. Find the co-ordinates of point B.

Section Formula

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Answer

Let coordinates of B be (a, b).

Point P(-4, 1) divides the line segment joining A(2, -2) and B(a, b) in the ratio 3 : 5.

The point P(-4, 1) divides the line segment joining the points A(2, -2) and B in the ratio 3 : 5. Find the co-ordinates of point B. Reflection, RSA Mathematics Solutions ICSE Class 10.

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 :

⇒(−4,1)=(3a+5(2)3+5,3b+5(−2)3+5)⇒(−4,1)=(3a+108,3b−108)⇒−4=(3a+108) and 1=(3b−108)⇒−32=3a+10 and 8=3b−10⇒−32−10=3a and 8+10=3b⇒−42=3a and 18=3b⇒a=−423 and b=183⇒a=−14 and b=6.\Rightarrow (-4, 1) = \Big(\dfrac{3a + 5(2)}{3 + 5}, \dfrac{3b + 5(-2)}{3 + 5}\Big) \\[1em] \Rightarrow (-4, 1) = \Big(\dfrac{3a + 10}{8}, \dfrac{3b - 10}{8}\Big) \\[1em] \Rightarrow -4 = \Big(\dfrac{3a + 10}{8}\Big) \text{ and } 1 = \Big(\dfrac{3b - 10}{8}\Big) \\[1em] \Rightarrow -32 = 3a + 10 \text{ and } 8 = 3b - 10 \\[1em] \Rightarrow -32 - 10 = 3a \text{ and } 8 + 10 = 3b \\[1em] \Rightarrow -42 = 3a \text{ and } 18 = 3b \\[1em] \Rightarrow a = \dfrac{-42}{3} \text{ and } b = \dfrac{18}{3} \\[1em] \Rightarrow a = -14 \text{ and }b = 6 .

B = (a, b) = (-14, 6).

Hence, coordinates of B(-14, 6).

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