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Mathematics

If the slope of the line joining P(k, 3) and Q(8, –6) is 34\dfrac{-3}{4}, find the value of k.

Straight Line Eq

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Answer

Given,

P(k, 3) and Q(8, –6)

Slope = 34\dfrac{-3}{4}

By formula,

Slope (m) = y2y1x2x1\dfrac{y2 - y1}{x2 - x1}

Substituting values we get,

Slope of PQ=638k34=638k34=98k3(8k)=4(9)243k=363k=36243k=12k=123k=4.\Rightarrow \text{Slope of PQ} = \dfrac{-6 - 3}{8 - k} \\[1em] \Rightarrow -\dfrac{3}{4} = \dfrac{-6 - 3}{8 - k} \\[1em] \Rightarrow -\dfrac{3}{4} = \dfrac{-9}{8 - k} \\[1em] \Rightarrow 3(8 - k) = 4(9) \\[1em] \Rightarrow 24 -3k = 36 \\[1em] \Rightarrow -3k = 36 - 24 \\[1em] \Rightarrow -3k = 12 \\[1em] \Rightarrow k = \dfrac{12}{-3} \\[1em] \Rightarrow k = -4.

Hence, k = -4.

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