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 $B,C$ and $D$ 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. $26$
• B. $25$
• C. $32$
• D. $27$

Answer 1

The correct option is A $26$

Let the equation of line From point $(-5,-4)$

$\frac{x+5}{\cos \theta }=\frac{y+4}{\sin \theta }=r$

Let $AB=r_1$

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

Since , the point B lies on $x+3y+2=0$

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

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

Let $AC=r_2$

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

Since , the point C lies on $2x+y+4=0$

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

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

Let $AD=r_3$

Then coordinates of $D(r_3\cos \theta -5,r_3\sin \theta -4)$

Since , the point C lies on $-y-5=0$

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

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

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

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

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

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

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

So the equation of required line is

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

$\Rightarrow 2x+3y+23=0$----(1)

Given equation is $2x+ay+b=0$ is equal to (1)

ON comaparing both eq

$a=3,b=23$

$2a+b=6+23=29$