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Mathematics

Find the value of p for which the points (-5, 1), (1, p) and (4, -2) are collinear.

Section Formula

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Answer

Since, the points A, B are C are collinear then the point A(-5, 1) divides BC in the ratio of m1 : m2.

For x-coordinate using section formula,

x=m1x2+m2x1m1+m2⇒−5=m1×4+m2×1m1+m2⇒−5=4m1+m2m1+m2⇒−5m1−5m2=4m1+m2⇒−5m1−4m1=m2+5m2⇒−9m1=6m2⇒m1m2=−69=−23….[Eq 1]x = \dfrac{m1x2 + m2x1}{m1 + m2} \\[1em] \Rightarrow -5 = \dfrac{m1 \times 4 + m2 \times 1}{m1 + m2} \\[1em] \Rightarrow -5 = \dfrac{4m1 + m2}{m1 + m2} \\[1em] \Rightarrow -5m1 - 5m2 = 4m1 + m2 \\[1em] \Rightarrow -5m1 - 4m1 = m2 + 5m2 \\[1em] \Rightarrow -9m1 = 6m2 \\[1em] \Rightarrow \dfrac{m1}{m2} = -\dfrac{6}{9} = -\dfrac{2}{3} \qquad \text{….[Eq 1]}

For y-coordinate using section formula,

y=m1y2+m2y1m1+m2⇒1=m1×(−2)+m2×pm1+m2⇒1=−2m1+m2pm1+m2⇒m1+m2=−2m1+m2p⇒m1+2m1=m2p−m2⇒3m1=m2(p−1)⇒m1m2=p−13….[Eq 2]y = \dfrac{m1y2 + m2y1}{m1 + m2} \\[1em] \Rightarrow 1 = \dfrac{m1 \times (-2) + m2 \times p}{m1 + m2} \\[1em] \Rightarrow 1 = \dfrac{-2m1 + m2p}{m1 + m2} \\[1em] \Rightarrow m1 + m2 = -2m1 + m2p \\[1em] \Rightarrow m1 + 2m1 = m2p - m2 \\[1em] \Rightarrow 3m1 = m2(p - 1) \\[1em] \Rightarrow \dfrac{m1}{m2} = \dfrac{p - 1}{3} \qquad \text{….[Eq 2]}

Comparing Eq 1 and 2,

⇒−23=p−13⇒−23=p−13⇒−2×3=3(p−1)⇒−6=3p−3⇒3p=−3⇒p=−1.\Rightarrow -\dfrac{2}{3} = \dfrac{p - 1}{3} \\[1em] \Rightarrow -\dfrac{2}{3} = \dfrac{p - 1}{3} \\[1em] \Rightarrow -2 \times 3 = 3(p - 1) \\[1em] \Rightarrow -6 = 3p - 3 \\[1em] \Rightarrow 3p = -3 \\[1em] \Rightarrow p = -1.

Hence, the value of p = -1.

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