Question

Prove that the following sets of three lines meet in a point.
$3x+4y+6=0,6x+5y+9=0$, and $3x+3y+5=0$.

Answer 1

$3x+4y+6=\mathrm{0......}\left(i\right)\Rightarrow x=\frac{-6-4y}{3}6x+5y+9=\mathrm{0.......}\left(ii\right)6\left(\frac{-6-4y}{3}\right)+5y+9=0-12-8y+5y+9=0-3y-3=0-3y=3\Rightarrow y=-1x=\frac{-6-4(-1)}{3}\Rightarrow x=-\frac{2}{3}$

So the point of intersection of $\left(i\right)$ and $\left(ii\right)$ is $(-\frac{2}{3},-1)$

$3x+3y+5=\mathrm{0......}\left(iii\right)$

Substituting the point of intersection in $\left(iii\right)$

$3(-\frac{2}{3})+3(-1)+5=-2-3+5=0$

$\left(iii\right)$ also satisfies the point of intersection of $\left(i\right)$ and $\left(ii\right)$ , so the given set of lines meet in a point .