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Find the coordinates of the point which is three-fourth of the way from A(3, 1) to B(-2, 5).

Section Formula

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Answer

Let P(x, y) be the required point. Then according to question,

Find the coordinates of the point which is three-fourth of the way from A(3, 1) to B(-2, 5). Section Formula, ML Aggarwal Understanding Mathematics Solutions ICSE Class 10.

AP = 34\dfrac{3}{4}AB

∴APAB=34⇒APAP+PB=34⇒4AP=3AP+3PB⇒4AP−3AP=3PB⇒AP=3PB⇒APPB=31∴m1=3,m2=1.\therefore \dfrac{AP}{AB} = \dfrac{3}{4} \\[1em] \Rightarrow \dfrac{AP}{AP + PB} = \dfrac{3}{4} \\[1em] \Rightarrow 4AP = 3AP + 3PB \\[1em] \Rightarrow 4AP - 3AP = 3PB \\[1em] \Rightarrow AP = 3PB \\[1em] \Rightarrow \dfrac{AP}{PB} = \dfrac{3}{1} \\[1em] \therefore m1 = 3, m2 = 1.

∴ By section formula coordinates of P are

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

Putting the values from question in above formula we get,

⇒(3×(−2)+1×33+1,3×5+1×13+1)=(−6+34,15+14)=(−34,164)=(−34,4).\Rightarrow \Big(\dfrac{3 \times (-2) + 1 \times 3}{3 + 1}, \dfrac{3 \times 5 + 1 \times 1}{3 + 1}\Big) \\[1em] = \Big(\dfrac{-6 + 3}{4}, \dfrac{15 + 1}{4}\Big) \\[1em] = \Big(-\dfrac{3}{4}, \dfrac{16}{4}\Big) \\[1em] = \Big(-\dfrac{3}{4}, 4\Big).

Hence, the coordinates of P are (−34,4).(-\dfrac{3}{4}, 4).

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