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Find a point P on the line segment joining A(14,-5) and (-4,4), which is twice as far from A as from B

Section Formula

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Answer

Let point P be (x, y).

P(1, -2) is a point on the line segment A(3, -6) and B(x, y) such that AP : PB is equal to 2 : 3. Find the co-ordinates of B. Reflection, RSA Mathematics Solutions ICSE Class 10.

Since, point P is twice as far from A as from B.

⇒ AP = 2BP

⇒ APBP=21\dfrac{AP}{BP} = \dfrac{2}{1}

⇒ AP : BP = 2 : 1.

⇒ m1 : m2 = AP : PB = 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 :

⇒(x,y)=(2×−4+1×142+1,2×4+1×−52+1)⇒(x,y)=(−8+143,8−53)⇒(x,y)=(63,33)⇒(x,y)=(2,1).\Rightarrow (x, y) = \Big(\dfrac{2 \times -4+ 1 \times 14}{2 + 1}, \dfrac{2 \times 4 + 1 \times -5}{2 + 1}\Big) \\[1em] \Rightarrow (x, y) = \Big(\dfrac{-8 + 14}{3}, \dfrac{8 - 5}{3}\Big) \\[1em] \Rightarrow (x, y) = \Big(\dfrac{6}{3}, \dfrac{3}{3}\Big) \\[1em] \Rightarrow (x, y) = (2, 1).

Hence, the coordinates of P are (2, 1).

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