The joint equation of pair of lines passing through point of intersection of lines $2 x^{2}-x y-15 y^{2}-7…
The joint equation of pair of lines passing through point of intersection of lines
$2 x^{2}-x y-15 y^{2}-7 x+32 y-9=0$ and parallel to co-ordinate axes is
$x y-x-2 y+2=0$
$x y+x+2 y-2=0$
$x y+x+2 y+2=0$
$x y-x-2 y-2=0$
Solution
Let $\phi=2 x^{2}-x y-15 y^{2}-7 x+32 y-9=0$ ...(1)
$\quad \frac{d \phi}{d x}=4 x-y-7=0 \Rightarrow 4 x-y-7=0$ ...(2)
$\quad \frac{d \phi}{d y}=-x-30 y+32=0 \Rightarrow x+30 y-32=0$ ...(3)
Solving equation $(2) \&(3)$ we get $x=2, y=1$
$\therefore(2,1)$ is the point of intersection of given lines.
Lines passing through $(2,1)$ and parallel to co-ordinate axes are $x=2$ and $y=1$.
Hence required equation is
$\quad(x-2)(y-1)=0 \Rightarrow x y-2 y-x+2=0$
Note : Point of intersection can also be calculated as follows :
$2 x^{2}-x y-15 y^{2}-7 x+32 y-9=0 \text { gives } a=2, h=\frac{-1}{2}, b=-15, g=\frac{-7}{2}, f=16, c=-9$
$\begin{aligned} \text { Point of intersection } &=\left(\frac{\mathrm{bg}-\mathrm{hf}}{\mathrm{h}^{2}-\mathrm{ab}}, \frac{\mathrm{af}-\mathrm{gh}}{\mathrm{h}^{2}-\mathrm{ab}}\right) \\ & \equiv\left[\frac{(-15)\left(-\frac{7}{2}\right)-\left(-\frac{1}{2}\right)(16)}{\left(-\frac{1}{2}\right)^{2}-(2)(-15)}, \frac{(2)(16)-\left(-\frac{7}{2}\right)\left(-\frac{1}{2}\right)}{\left(-\frac{1}{2}\right)^{2}-(2)(-15)}\right], \\ & \equiv\left[\frac{\frac{105}{2}+\frac{16}{2}}{\frac{1}{4}+30}, \frac{32-\frac{7}{4}}{\frac{1}{4}+30}\right] \equiv\left[\frac{121}{2} \times \frac{4}{121}, \frac{121}{4} \times \frac{4}{121}\right] \\ & \equiv(2,1) \end{aligned}$