Question

The equation of the line which is equidistant from the lines $y=-\frac{13}{2}$ and $y=\frac{17}{2}$ is :-

• A. $y=2$
• B. $y=\frac{1}{2}$
• C. $y=-1$
• D. $y=1$

Answer 1

The correct option is D $y=1$

Let any point '$P$' on the line be $(x,y)$
$PA=PB$......(given)

$P{A}^2=P{B}^2$

Since the given lines are $y=-\frac{13}{2}$ and $y=\frac{17}{2}$

${(y-(-\frac{13}{2}))}^2={(y-\frac{17}{2})}^2\Rightarrow {(y+\frac{13}{2})}^2={(y-\frac{17}{2})}^2\Rightarrow y^2+\frac{169}{4}+13y=y^2+\frac{289}{4}-17y\Rightarrow y^2-y^2+\frac{169}{4}-\frac{289}{4}+13y+17y=0\Rightarrow -30y=-\frac{120}{4}\Rightarrow -30y=-30\Rightarrow y=1$

Hence, the equation of the line is $y=1$.