Question

Joint equation of pair of lines through $(3,-2)$ and parallel to $x^2-4xy+3y^2=0$ is

• A. $x^2+3y^2-4xy-14x+24y+45=0$
• B. $x^2+3y^2+4xy-14x+24y+45=0$
• C. $x^2+3y^2+4xy-14x+24y-45=0$
• D. $x^2+3y^2+4xy-14x-24y-45=0$

Answer 1

The correct option is A $x^2+3y^2-4xy-14x+24y+45=0$

Given equation of line is $x^2-4xy+3y^2=0$
$\therefore m_1+m_2=\frac{4}{3}$ and $m_1{m}_2=\frac{1}{3}$
On solving these equations, we get
$m_1=1,m_2=\frac{1}{3}$
Let the lines parallel to given line are $y=m_{1}x+c_1$ and $y=m_{2}x+c_2$
$\therefore y=\frac{1}{3}x+c_1$ and $y=x+c_2$
Also, these lines passes through the point $(3,-2)$
$\therefore -2=\frac{1}{3}\times 3+c_1$
$\Rightarrow c_1=-3$
and $-2=1\times 3+c_2$
$\Rightarrow c_2=-5$
$\therefore$ Required equation of pair of lines is $(3y-x+9)(y-x+5)=0$
$\Rightarrow x^2+3y^2-4xy-14x+24y+45=0$