Question

In a triangle $ABC$, if $A$ is $(2,-1)$ and $7x-10y+1=0$ and $3x-2y+5=0$ are the equations of an altitude and an angle bisector, respectively, drawn from $B$, then the equation of $BC$ is

• A. $x+y+1=0$
• B. $5x+y+17=0$
• C. $4x+9y+30=0$
• D. $x-5y-7=0$

Answer 1

Answer: (B)

$BD$ and $BE$ intersect at $B$. The coordinates of $B$ are $(-3,-2)$
$m_{AB}=\frac{1}{5}$

or $\tan \theta =\left|\frac{\frac{3}{2}-\frac{1}{5}}{1+\frac{3}{10}}\right|=\left|\frac{\frac{3}{2}-m}{1+\frac{3m}{2}}\right|$
or $1=\left|\frac{3-2m}{2+3m}\right|$
or $\pm 1=\frac{3-2m}{2+3m}$
i.e., $m=\frac{1}{5}$ (rejected) or $m=-5$
The equation of $BC$ is
$y+2=-5(x+3)$
or $5x+y+17=0$