Question

If a line $2x+ay+b=0$ through $A(-5,-4)$ meets the lines $x+3y+2=0,2x+y+4=0,$ and $x-y-5=0$ at the points $B,C,\&D,$ respectively, which satisfy ${\left(\frac{15}{AB}\right)}^2+{\left(\frac{10}{AC}\right)}^2={\left(\frac{6}{AD}\right)}^2$. Then the value of $2a+b$ is

• A. $29$
• B. $25$
• C. $32$
• D. $27$

Answer 1

The correct option is A $29$

Let the eq through $A(-5,-4)$

$\frac{x+5}{\cos \theta }=\frac{y+4}{\sin \theta }=r$-----(1)

Let $AB=r_1$

SO coordinates of $B(r_1\cos \theta -5,r_1\sin \theta -4)$

Since B lies in $x+3y+2=0$

$\Rightarrow r_1(3\sin \theta +\cos \theta )=15$

$\Rightarrow \frac{15}{r_1}=3\sin \theta +\cos \theta$------(2)

Let $AC=r_2$

SO coordinates of $C(r_2\cos \theta -5,r_2\sin \theta -4)$

Since C lies in $2x+y+4=0$

$\Rightarrow r_2(\sin \theta +2\cos \theta )=10$

$\Rightarrow \frac{10}{r_2}=\sin \theta +2\cos \theta$------(3)

Let $AD=r_3$

SO coordinates of $D(r_2\cos \theta -5,r_2\sin \theta -4)$

Since D lies in $x-y-5=0$

$\Rightarrow r_3(-\sin \theta +\cos \theta )=6$

$\Rightarrow \frac{6}{r_3}=-\sin \theta +\cos \theta$------(3)

$\Rightarrow {\left(\frac{15}{AB}\right)}^2+{\left(\frac{10}{AC}\right)}^2={\left(\frac{6}{AD}\right)}^2$

$\Rightarrow {\left(\frac{15}{r_1}\right)}^2+{\left(\frac{10}{r_2}\right)}^2={\left(\frac{6}{r_3}\right)}^2$

$\Rightarrow (3\sin \theta +\cos \theta {)}^2+(\sin \theta +2\cos \theta {)}^2=(-\sin \theta +\cos \theta {)}^2$

$\Rightarrow (3\sin \theta +2\cos \theta {)}^2=0$

$\Rightarrow \tan \theta =-\frac{2}{3}$

SO eq be

$y+5=-\frac{2}{3}(x+4)$

$2x+3y+23=0$

Comparing above eq with $2x+ay+b=0$

$a=3,b=23$

$2a+b=2\left(3\right)+23=29$