Question

Show that the equation $2x^2-xy-3y^2-6x+19y-20=0$ represent a pair of intersecting lines. Show that the angle between them is $ta{n}^{-1}\left(5\right)$.

Answer 1

$2x^2-xy-3y^2-6x+19y-20=D$

$h^2>ab\Rightarrow {\left(\frac{-1}{2}\right)}^2>2\times (-3)$

$\Rightarrow \frac{1}{4}>-6=tan\theta =|\frac{2\sqrt{h^2-ab}}{a+b}|$

$=|\frac{2\frac{1}{4}+6}{2-3}|=\frac{5}{2}\times 2$

$=5$