Question 31

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Jambmaths question: 

$3y=4x-1$and $ky=x+3$are equation of two straight lines. If the two lines are perpendicular to each other, find k 

Option A: 

$-\tfrac{4}{3}$

Option B: 

$-\tfrac{3}{4}$

Option C: 

$tfrac{3}{4}$

Option D: 

$\tfrac{4}{3}$

Jamb Maths Solution: 

$\begin{align}  & {{l}_{1}}:3y=4x-1 \\ & y=\tfrac{4}{3}x-\tfrac{1}{3}\text{      (}y=mx+c\text{)} \\ & {{m}_{1}}=\tfrac{4}{3} \\ & {{l}_{2}}:ky=x+3 \\ & y=\tfrac{x}{k}+\tfrac{3}{k}\text{    }(y=mx+c) \\ & {{m}_{2}}=\tfrac{1}{k} \\ & \text{For two lines to be }\bot \text{ar} \\ & \text{ }{{m}_{1}}{{m}_{2}}=-1 \\ & \frac{4}{3}\times \frac{1}{k}=-1 \\ & k=-\frac{4}{3} \\\end{align}$

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