Question

Equation of a straight line passing through the point $(4,5)$ and equally inclined to the lines $3x=4y+7$ and $5y=12x+6$ is

• A. $9x-7y=1$
• B. $9x+7y=71$
• C. $7x+9y=73$
• D. $7x-9y+17=0$

Answer 1

Answer: (A), (C)

$7x+9y=73$

The required lines are parallel to the angle bisectors of the given lines. The angle bisectors of the given lines are
$\frac{3x-4y-7}{\sqrt{9+16}}=\pm \left(\frac{12x-5y+6}{\sqrt{144+25}}\right)$
$\Rightarrow 13(3x-4y-7)$ = $\pm 5(12x-5y+6)$
$\Rightarrow 21x+27y+121=0$ or $99x-77y-61=0$
Equations of the lines passing through (4, 5) and parallel to these lines are:
$21(x-4)+27(y-5)=0$
and $99(x-4)-77(y-5)=0$
i.e., the required lines are
$7x+9y-73=0$
and $9x-7y-1=0$