Question

Find k, if one of the lines given by $k{x}^2+10xy+8y^2=0$ is perpendicular to the line $2x-y=5$.

Answer 1

The slope of $2x-y=5$ is $2$.

So the slope of a perpendicular line is $\frac{-1}{2}$

$k{x}^2+10xy+8y^2=0$ is supposed to be the intersection of $2$ straight lines so, lets solve it for $y.$

$\Rightarrow$ $y=\frac{-10x\pm \sqrt{100x^2-32k{x}^2}}{16}$

$=\frac{-10x\pm x\sqrt{100-68k}}{16}$

$=\frac{x(-10\pm \sqrt{100-32k})}{16}$

So the slope of the two lines is $\frac{(-10\pm \sqrt{100-32k})}{16}$

$\therefore$ $\frac{(-10\pm \sqrt{100-32k})}{16}=\frac{-1}{2}$

$\Rightarrow$ $-20\pm 2\sqrt{100-32k}=-16$

$\Rightarrow$ $\pm 2\sqrt{100-32k}=4$

$\Rightarrow$ $\pm \sqrt{100-32k}=2$

$\Rightarrow$ $100-32k=4$ [ Squaring both sides ]

$\Rightarrow$ $32k=96$

$\therefore$ $k=3$