Question

Find the equation of the line passing through the point $(0,-2)$ and the point of intersection of the lines $4x+3y=1$ and $3x-y+9=0$.

Answer 1

Given that,

$4x+3y=1⟶\left(1\right)$

$3x-y+9=0⟶\left(2\right)$

Solving the equations simultaneously, we get
$4x+3y=1\times \mathrm{1....}\left(i\right)3(3x-y)=-9\times \mathrm{3......}\left(ii\right)\left[\text{Multiply by}3\right]\text{Adding (i) and (ii)}13x=-26x=-2y=-3\text{Hence point is}(-2,-3)$

$\therefore$ Slope of the passing through the point $(0,-2)and(-2,-3)$

$\Rightarrow m=\frac{-3+2}{-2-0}=\frac{-1}{-2}$ $\Rightarrow m=\frac{1}{2}$ $[\because m=\frac{y_2-y_1}{x_2-x_1}]$

So, the equation of line is given as,

$y-y_1=m(x-x_1)$

$\Rightarrow$ $(y+2)=\frac{1}{2}(x-0)$

$\Rightarrow$$2y+4=x$

$\Rightarrow$$x-2y+4=0$