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If A(2, –5), B(–2, 5), C(k, 3) and D(1, 1) be four points such that AB and CD are perpendicular to each other, find the value of k.

Straight Line Eq

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Answer

By using slope formula,

m = y2−y1x2−x1\dfrac{y2 - y1}{x2 - x1}

Given, points A(2, –5), B(–2, 5)

Substituting values we get,

mAB=5−(−5)−2−2=5+5−4=−104=−52m_{AB} = \dfrac{5 - (-5)}{-2 - 2} = \dfrac{5 + 5}{-4} = -\dfrac{10}{4} = -\dfrac{5}{2}

Given, points C(k, 3) and D(1, 1)

Substituting values we get,

mCD=1−31−k=−21−km_{CD} = \dfrac{1 - 3}{1 - k} = \dfrac{-2}{1 - k}

Since the lines are perpendicular the product of the gradients is equal to -1:

⇒(−52)×(−21−k)=−1⇒(102(1−k))=−1⇒(51−k)=−1⇒5=−1(1−k)⇒5=−1+k⇒k=5+1⇒k=6.\Rightarrow \Big(-\dfrac{5}{2}\Big) \times \Big(\dfrac{-2}{1 - k}\Big) = -1 \\[1em] \Rightarrow \Big(\dfrac{10}{2(1 - k)}\Big) = -1 \\[1em] \Rightarrow \Big(\dfrac{5}{1 - k}\Big) = -1 \\[1em] \Rightarrow 5 = -1(1 - k) \\[1em] \Rightarrow 5 = -1 + k \\[1em] \Rightarrow k = 5 + 1 \\[1em] \Rightarrow k = 6.

Hence, value of k = 6.

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