Question

Find the equation to the straight line passing through the point $(3,2)$ and the point of intersection of the lines $2x+3y=1$ and $3x-4y=6$.

Answer 1

$3x-4y=\mathrm{6.......}\left(i\right)2x+3y=12x=1-3yx=\frac{1-3y}{2}$

Substituting $x$ in $\left(i\right)$, we get

$3\left(\frac{1-3y}{2}\right)-4y=63-9y-8y=12-17y=9y=-\frac{9}{17}$

Now, $x=\frac{1-3y}{2}$

$\Rightarrow x=\frac{1-3(-\frac{9}{17})}{2}=\frac{1+\frac{27}{17}}{2}=\frac{22}{17}$

So the point of intersection is $P(\frac{22}{17},-\frac{9}{17})$

Equation of line joining $(3,2)$ and $P$ is

$y-2=\left(\frac{-\frac{9}{17}-2}{\frac{22}{17}-3}\right)(x-3)y-2=\left(\frac{-\frac{43}{17}}{-\frac{29}{17}}\right)(x-3)y-2=\frac{43}{29}(x-3)$

$43x-29y=71$.