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In what ratio is the join of (4, 3) and (2, -6) divided by the x-axis. Also, find the co-ordinates of the point of intersection.

Section Formula

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Answer

Let the point on x-axis be (x, 0) and required ratio be k : 1.

By formula,

y=m1y2+m2y1m1+m2y = \dfrac{m1y2 + m2y1}{m1 + m2}

Substituting values we get,

0=k×6+1×3k+10=6k+36k=3k=36=12.\Rightarrow 0 = \dfrac{k \times -6 + 1 \times 3}{k + 1} \\[1em] \Rightarrow 0 = -6k + 3 \\[1em] \Rightarrow 6k = 3 \\[1em] \Rightarrow k = \dfrac{3}{6} = \dfrac{1}{2}.

k : 1 = 12:1=1:2.\dfrac{1}{2} : 1 = 1 : 2.

x=m1x2+m2x1m1+m2x = \dfrac{m1x2 + m2x1}{m1 + m2}

Substituting values we get,

x=1×2+2×41+2x=2+83x=103.\Rightarrow x = \dfrac{1 \times 2 + 2 \times 4}{1 + 2} \\[1em] \Rightarrow x = \dfrac{2 + 8}{3} \\[1em] \Rightarrow x = \dfrac{10}{3}.

P = (x, 0) = (103,0).\Big(\dfrac{10}{3}, 0\Big).

Hence, ratio = 1 : 2 and co-ordinates of point of intersection = (103,0).\Big(\dfrac{10}{3}, 0\Big).

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