Question

Two equal sides of an isosceles triangle are given by $7x-y+3=0$ and $x+y-3=0$. If the slope $m$ of the third side is an integer, then $m=$

• A. 3
• B. 1
• C. -1
• D. -3

Answer 1

Answer: (D)

Given equation of sides are

$7x-y+3=0\dots (i)$
$x+y-3=0\dots (ii)$
Equation of angle bisectors of Eqs. (i) and (ii) are
$\frac{7x-y+3}{\sqrt{49+1}}=\pm \frac{x+y-3}{\sqrt{1+1}}$
$\frac{7x-y+3}{\sqrt{50}}=\pm \frac{x+y-3}{\sqrt{2}}$
$\frac{7x-y+3}{5}=\pm \frac{x+y-3}{1}$
$\therefore \frac{7x-y+3}{5}=(x+y-3)$
or $\frac{7x-y+3}{5}=-(x+y-3)$
or $7x-y+3=5x+5y-15$
$7x-y+3=-5x-5y+15$
$\Rightarrow 2x-6y+12=0$
or $12x+4y-12=0$
$\Rightarrow x-3y+6=0$ or $3x+y-3=0$
Slope $=\frac{1}{3}$, Slope $=-3$ (integer)
Here, third side is parallel to one angle bisector. $\therefore$ Required slope of third side is $-3$.