Question

Show that the following equation represents a pair of straight lines. Find the points of intersection and the acute angle between them.
(i) $3x^2+7xy+2y^2+5x+5y+2=0$
(ii) $2x^2-13xy-7y^2+x+23y-6=0$.

Answer 1

for a general eq $a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$ will represents pair of straight line if:

$a,b,h\ne 0$ and $h^2-ab>or=0$

1. for $3x^2+7xy+2y^2+5x+5y+2=0$ $a=3,h=7/2,b=2$
   $h^2-ab=49/4-6=25/4>0$ (represents straight line)
2. for $2x^2-13xy-7y^2+x+23y-6=0$

$a=2,h=-13/2,b=-7$

$h^2-ab=169/4+14=225/4>0$ (represents straight line)

Points of intersection is $\alpha =\frac{hf-bg}{ab-b^2}$ and $\beta =\frac{hg-af}{ab-h^2}$

after substituting values of $a,b,h,g,f$ we will get the points of intersection for the given lines.

acute angle $\theta =co{s}^{-1}\frac{|a+b|}{\sqrt{(a-b{)}^2+4h^2}}$

again, after putting the values of $a,b,h$ will get the acute angle.