Question 33

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Jambmaths question: 

Find the equation of a line perpendicular to the line 2y = 5x + 4 which passes (4, 2).

Option A: 

$5y-2x+18=0$

Option B: 

$5y+2x-2=0$

Option C: 

$5y+2x-18=0$

Option D: 

$5y-2x-18=0$

Jamb Maths Solution: 

$\begin{align}  & {{l}_{1}}:2y=5x+4 \\ & y=\tfrac{5}{2}x+2\text{     }\!\!\{\!\!\text{ }y=mx+c\} \\ & {{m}_{1}}=\frac{5}{2}\text{   (slope of }{{l}_{1}}) \\ & \text{For lines to be perpendicular} \\ & {{m}_{1}}{{m}_{2}}=-1 \\ & {{m}_{2}}=-\frac{1}{{{m}_{1}}}=-\frac{1}{{\scriptstyle{}^{5}\!\!\diagup\!\!{}_{2}\;}}=-\frac{2}{5} \\ & \text{Using one point slope from equation} \\ & \text{The equation of the second line will be } \\ & y-{{y}_{1}}=m(x-{{x}_{1}}) \\ & y-2=-\tfrac{2}{5}(x-4) \\ & 5y-10=-2x+8 \\ & 5y+2x-18=0 \\\end{align}$

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