Question 39

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Jambmaths question: 

Find the value of $\alpha $if the line $2y-\alpha x+4=0$is perpendicular to the line $y+\tfrac{1}{4}x-7=0$

Option A: 

–4

Option B: 

4

Option C: 

8

Option D: 

–8

Jamb Maths Solution: 

$\begin{align}  & \text{Equation of a line is given as } \\ & y=mx+c \\ & \text{where }m=gradient,\text{ }c=\operatorname{int}ercept \\ & y+\tfrac{1}{4}x-7=0 \\ & y=-\tfrac{1}{4}x+7\text{      }\!\!\{\!\!\text{ }y=mx+c\} \\ & {{m}_{1}}=-\frac{1}{4} \\ & \text{Also for the line }2y-\alpha x+4=0 \\ & 2y=\alpha x-4 \\ & y=\frac{1}{2}\alpha x-2\text{    }\{y=mx+c\} \\ & {{m}_{2}}=\frac{1}{2}\alpha  \\ & \text{For two lines to be perpendicular } \\ & {{m}_{1}}{{m}_{2}}=-1 \\ & -\frac{1}{4}\times \frac{\alpha }{2}=-1 \\ & \frac{\alpha }{2}=4 \\ & \alpha =8 \\\end{align}$

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