Question

Find the point of intersection and the angle between the lines given by the equation : $2x^2+3xy-2y^2+10x+5y+12=0$

• A. $(-\frac{11}{5},-\frac{2}{5}),{90}^o$
• B. $(\frac{11}{5},-\frac{2}{5}),{45}^o$
• C. $(-\frac{11}{5},\frac{2}{5}),{90}^o$
• D. $(\frac{11}{5},\frac{2}{5}),{60}^o$

Answer 1

The correct option is A $(-\frac{11}{5},-\frac{2}{5}),{90}^o$

Given lines,
$2x^2+3xy-2y^2+10x+5y+12=0$ and from general equation $a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$ when $h^2>ab$ ,we know that point of intersection is given by $(\frac{hf-bg}{ab-h^2},\frac{gh-af}{ab-h^2})$ ,By substituting values
We get point of intersection as $(-\frac{11}{5},-\frac{2}{5})$ and angle between the lines is given by $\tan \theta =\frac{2\sqrt{h^2-ab}}{a+b}$ and by substituting values ,we get angle between them is $\tan \theta =\infty ⟹\theta ={90}^0$