If the equation of the pair of straight lines passing through the point $(1,1)$ and perpendicular to the…
If the equation of the pair of straight lines passing through the point $(1,1)$ and perpendicular to the pair of lines $3 \mathrm{x}^2$ $+11 x y-4 y^2=0$ is $\mathrm{a}^2+2 \mathrm{~h} x y+\mathrm{b} y^2+2 \mathrm{~g} x+2 \mathrm{f} y+12=0$, then $2(a-h+b-g+f-12)=$
$0$
$-7$
$-19$
$13$
Solution
$3 x^2+11 x y-4 y^2=0 \Rightarrow(3 x-y)(x+4 y)=0$
Equation of lines are
$3 x-y=0 \ldots$ (i); $x+4 y=0 \ldots$ (ii)
$\therefore \quad m_1=3, m_2=\frac{-1}{4}$
So lines perpendicular to (i) and (ii) will have slope $\frac{-1}{2}, 4$
Also the lines passes through $(1,1)$
$\therefore y-1=-\frac{1}{3}(x-1) \Rightarrow x+3 y=4$ ....(iii)
and $y-1=4(x-1) \Rightarrow 4 x-y=3$ ....(iv)
Combined equation of (iii) and (iv) is
$(x+3 y-4)(4 x-y-3)=0$
$\Rightarrow 4 x^2-3 y^2+11 x y-19 x-5 y+12=0$
$\Rightarrow a=4, b=-3, h=\frac{11}{2}, g=\frac{-19}{2}, f=\frac{-5}{2}$
$2(a-h+b-g+f-12)=-19$