Find the equation to the straight line passing through the point $\left(3,2\right)$ and the intersection of the lines $2x + 3y = 1$ and $3x - 4y = 6$.
Comparing $\;$ $2x + 3y = 1$ $\;\;\; \cdots \; (1)$ $\;$ and $\;$ $3x - 4y = 6$ $\;\;\; \cdots \; (2)$
with the standard equations $\;$ $a_1x + b_1 y + c_1 = 0$ $\;$ and $\;$ $a_2 x + b_2 y + c_2 = 0$ respectively gives
$a_1 = 2$, $\;$ $b_1 = 3$, $\;$ $c_1 = -1$, $\;$ $a_2 = 3$, $\;$ $b_2 = -4$, $\;$ $c_2 = -6$
The coordinates point of intersection of equations $(1)$ and $(2)$ are
$x = \dfrac{b_1 c_2 - b_2 c_1}{a_1 b_2 - a_2 b_1}$ $\;$ and $\;$ $y = \dfrac{c_1 a_2 - c_2 a_1}{a_1 b_2 - a_2 b_1}$
i.e. $\;$ $x = \dfrac{3 \times \left(-6\right) + 4 \times \left(-1\right)}{2 \times \left(-4\right) - 3 \times 3}$ $\;$ and $\;$ $y = \dfrac{-1 \times 3 + 6 \times 2}{2 \times \left(-4\right) - 3 \times 3}$
i.e. $\;$ $x = \dfrac{-22}{-17}$, $\;$ $y = \dfrac{9}{-17}$
i.e. $\;$ The point of intersection is $\;$ $= \left(x, y\right) = \left(\dfrac{22}{17},\dfrac{-9}{17}\right)$
The required equation passes through the points $\left(x_1, y_1\right) = \left(3, 2\right)$ and $\left(x_2, y_2\right) = \left(\dfrac{22}{17},\dfrac{-9}{17}\right)$
Slope of the required line $= m = \dfrac{y_2 - y_1}{x_2 - x_1}$
i.e. $\;$ $m = \dfrac{\dfrac{-9}{17} - 2}{\dfrac{22}{17} - 3} = \dfrac{43}{29}$
Equation of the required line is of the form $\;$ $y - y_1 = m \left(x - x_1\right)$
i.e. $\;$ $y - 2 = \dfrac{43}{29} \left(x - 3\right)$
i.e. $\;$ $29 y - 58 = 43 x - 129$
i.e. $\;$ $43 x - 29y - 71 = 0$