Question 32

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Jambmaths question: 

Find the gradient of a line which is perpendicular to the line with the equation $3x+2y+1=0$

Option A: 

$\tfrac{3}{2}$

Option B: 

$\tfrac{2}{3}$

Option C: 

$-\tfrac{2}{3}$

Option D: 

$-\tfrac{3}{2}$

Jamb Maths Solution: 

$\begin{align}  & {{l}_{1}}:3x+2y+1=0 \\ & 2y=-3x-1 \\ & y=-\tfrac{3}{2}x-\tfrac{1}{2}\text{   (}y=mx+c) \\ & m=\text{ slope or gradient} \\ & \text{slope of }{{l}_{1}}={{m}_{1}}=-\tfrac{3}{2} \\ & \text{For two lines to be }\bot \text{ar} \\ & {{m}_{1}}{{m}_{2}}=-1 \\ & {{m}_{2}}=-\frac{1}{{{m}_{1}}}=-\frac{1}{-\tfrac{3}{2}} \\ & {{m}_{2}}=\frac{2}{3} \\ & \text{The slope of the line that is pendicular to }3x+2y+1=0\text{ is }\frac{2}{3} \\\end{align}$

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