Question

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

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

Answer 1

Answer: (D)

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