Find the equation to the straight line passing through the intersection of the lines $x - 2y -3 = 0$ and $x + 3y - 6 = 0$ and parallel to the line $3x + 4y = 7$
Equation of a line passing through the intersection of the lines
$x - 2y -3 = 0$ $\;\;\; \cdots \; (1)$ $\;$ and $\;$ $x + 3y - 6 = 0$ $\;\;\; \cdots \; (2)$ is
$\left(x - 2y -3\right) + \lambda \left(x + 3y - 6\right) = 0$ $\;\;\; \cdots \; (3)$ $\;$ where $\lambda$ is a scalar.
i.e. $\;$ $\left(1 + \lambda\right)x + \left(-2 + 3 \lambda\right)y - \left(3 + 6 \lambda\right) = 0$ $\;\;\; \cdots \; (4)$
i.e. $\;$ $y = \left(\dfrac{1 + \lambda}{2 - 3 \lambda}\right) x + \dfrac{3 + 6 \lambda}{-2 + 3 \lambda}$
$\therefore \;$ Slope of equation $(3)$ is $= m_1 = \dfrac{1 + \lambda}{2 - 3 \lambda}$
Equation of given line is: $\;$ $3x + 4y = 7$
i.e. $\;$ $y = \dfrac{-3}{4} x + \dfrac{7}{4}$
$\therefore \;$ Slope of given line $= m_2 = \dfrac{-3}{4}$
Since the required line is parallel to the given line, $\;$ $m_1 = m_2$
i.e. $\;$ $\dfrac{1 + \lambda}{2 - 3 \lambda} = \dfrac{-3}{4}$
i.e. $\;$ $4 + 4 \lambda = -6 + 9 \lambda$
i.e. $\;$ $5 \lambda = 10$ $\implies$ $\lambda = 2$
Substituting the value of $\lambda$ in equation $(4)$ gives the equation of the required line as
$3x + 4y - 15 = 0$