Question 7

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Jambmaths question: 

What is the locus of points equidistant from the ax + by + c = 0 

Option A: 

A line ax + by + q = 0

Option B: 

A line axby + q = 0

Option C: 

A line bxay + q = 0

Option D: 

A line bx + ay + q = 0

Jamb Maths Solution: 

$\begin{align}  & \text{The line that will be equidistant from }ax+by+c=0\text{ must } \\ & \text{be perpendicular bisector of the line }ax+by+c=0. \\ & \text{For two lines to be perpendicular},{{m}_{1}}{{m}_{2}}=1. \\ & \text{That is the product of slopes of the two lines must be equal to}~\text{1} \\ & ax+by+c=0 \\ & by=-ax-c \\ & y=\frac{-a}{b}x-\frac{c}{b}\text{    }(y=mx+c) \\ & {{m}_{1}}=-\frac{a}{b} \\ & {{m}_{1}}{{m}_{2}}=-1\text{     (condition for perpendicularity)} \\ & {{m}_{2}}=-\frac{1}{{{m}_{1}}} \\ & {{m}_{2}}=-\frac{1}{-{\scriptstyle{}^{a}\!\!\diagup\!\!{}_{b}\;}}=\frac{b}{a} \\ & \text{Equation of the second line} \\ & y=\frac{b}{a}x+k\text{    (}y=mx+c) \\ & ay=bx+ak \\ & bx-ay+ak=0 \\ & \text{Let }ak=q \\ & bx-ay+q=0 \\ & \text{The equation of the second line is }bx-ay+q=0 \\\end{align}$

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