Question

A straight line through the point $A(-2,-3)$ cuts the line $x+3y=9$ and $x+y+1=0$ at $B$ and $C$ respectively. The equation of the line if $AB.AC=20$ is

• A. $x-y=1$
• B. $3x-y+3=0$
• C. $x-y=-1$
• D. $3x-y=3$

Answer 1

The correct option is A $x-y=1$

$\begin{array}{c}\frac{x+2}{\cos \theta }=\frac{y+3}{Sin\theta }=r\\ r-r_{1}andr=r_2\\ r_1\cos \theta -2+r_{2}Sin\theta -3+1=0\\ r_1\left(\cos \theta +Sin\theta \right)=4\\ r_2\left(\cos \theta +Sin\theta \right)=20\\ r_1{r}_2=20=\frac{80}{\left(\cos \theta +Sin\theta \right)\left(\cos \theta +3Sin\theta \right)}\\ \tan \theta =3or1\\ eq.ofliney+3=3\left(x+2\right)\\ y+3=\left(x+2\right)\end{array}$