Question

The combined equation of the pair of straight lines passing through the point of intersection of the pair of lines $x^{2}+4 x y+3 y^{2}-4 x-10 y+3=0$ and having slopes $\frac{1}{2}$ and $-\frac{1}{3}$ is

• A. $x^{2}-y^{2}-8 x-2 y+15=0$
• B. $x^{2}+7 x y+12 y^{2}-x-4 y=0$
• C. $x^{2}+7 x y+10 y^{2}-x-8 y-2=0$
• D. $x^{2}+x y-6 y^{2}-7 x-16 y+6=0$

Answer 1

Answer: (D)

Combined equation of the pair of straight line $x^{2}+4 x y+3 y^{2}-4 x-10 y+3=0$ is in the form of $a x^{2}+2 h x y+b y^{2}+2 g x+2 f y+c=0$
$\therefore \,a=1,2 h=4$
$ \Rightarrow \,h=2, \,b=3, \,g=-2,\, f=-5,\, c=3$
So, point of intersection of pair of straight line is
$\left(\frac{b g-f h}{h^{2}-a b}, \frac{a f-g h}{h^{2}-a b}\right)$
$=\left(\frac{-6+10}{4-3}, \frac{-5+4}{4-3}\right)=(4,-1)$
Equation of line having slope $\frac{1}{2}$ and passes through $(4,-1)$ is
$(y-(-1)) =\frac{1}{2}(x-4) $
$2(y+1) =x-4$
$\Rightarrow \, x-2 y-6=0\,\,\,\,\,\,\dots(i)$
And equation of line having slope $\frac{-1}{3}$ and passes through $(x,-1)$ is $y-(-1)=-\frac{1}{3}(x-4)$
$\Rightarrow \, 3 y+3=-x+4$
$\Rightarrow \, x+3 y-1=0$
Combined equation of pair of straight line of Eqs. (i) and (ii) is
$(x-2 y-6)(x+3 y-1)=0$
$\Rightarrow \, x^{2}+x y-6 y^{2}-7 x-16 y+6=0$