Find the coordinates of the point of intersection of the lines $2x - 3y = 1$ and $5y - x = 3$, and find the acute angle at which they cut each other.
Comparing $\;$ $2x - 3y = 1$ $\;\;\; \cdots \; (1)$ $\;$ and $\;$ $5y - x = 3$ $\;\;\; \cdots \; (2)$
with the standard equations $\;$ $a_1x + b_1 y + c_1 = 0$ $\;$ and $\;$ $a_2 x + b_2 y + c_2 = 0$ respectively gives
$a_1 = 2$, $\;$ $b_1 = -3$, $\;$ $c_1 = -1$, $\;$ $a_2 = 1$, $\;$ $b_2 = -5$, $\;$ $c_2 = 3$
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. $\;$ $x = \dfrac{-3 \times 3 + 5 \times \left(-1\right)}{2 \times \left(-5\right) - 1 \times \left(-3\right)}$ $\;$ and $\;$ $y = \dfrac{-1 \times 1 - 3 \times 2}{2 \times \left(-5\right) - 1 \times \left(-3\right)}$
i.e. $\;$ $x = \dfrac{-14}{-7}$, $\;$ $y = \dfrac{-7}{-7}$
i.e. $\;$ The point of intersection is $\;$ $= \left(x, y\right) = \left(2,1\right)$
Equation $(1)$ can be written as $\;$ $y = \dfrac{2}{3}x - \dfrac{1}{3}$ $\;\;\; \cdots \; (3)$
Equation $(2)$ can be written as $\;$ $y = \dfrac{1}{5}x + \dfrac{3}{5}$ $\;\;\; \cdots \; (4)$
Slope of equation $(3)$ is $= m_1 = \dfrac{2}{3}$
Slope of equation $(4)$ is $= m_2 = \dfrac{1}{5}$
Let $\theta$ be the angle between the lines given by equations $(1)$ and $(2)$.
Then $\;$ $\theta = \tan^{-1} \left(\dfrac{m_1 - m_2}{1 + m_1 m_2}\right)$
i.e. $\;$ $\theta = \tan^{-1} \left(\dfrac{\dfrac{2}{3} - \dfrac{1}{5}}{1 + \dfrac{2}{3} \times \dfrac{1}{5}}\right)$
i.e. $\;$ $\theta = \tan^{-1} \left(\dfrac{7}{17}\right)$