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

Answer 1

Answer: (A)

$x^2 + 3y^2 - 4xy - 14x + 24y + 45 = 0$

Answer 2

Given equation of line is $x^{2}-4 x y+3 y^{2}=0$ $\therefore m_{1}+m_{2}=\frac{4}{3}$ 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 $(3 y-x+9)(y-x+5)=0$ $\Rightarrow \,\, x^{2}+3 y^{2}-4 x y-14 x+24 y+45=0$