WAEC 2020 · Paper 2 · Q9

(a)
Two lines $L_1$ and $L_2$ pass through the point of intersection of $2y = x - 13$ and $3y + x + 12 = 0$ . $L_1$ passes through $P(-4, -7)$ and $L_2$ is perpendicular to $2x - 5y = 4$ . Find the acute angle between $L_1$ and $L_2$ (2 d.p.).

L₁ (blue) through P and the intersection; L₂ (red) perpendicular to 2x − 5y = 4 (grey).

Worked solution (try it first)

From $2y = x - 13$ , $x = 2y + 13$ .
Substitute: $3y + 2y + 13 + 12 = 0$ , so $y = -5$ and $x = 3$ .
$L_1$ through $(3, -5)$ and $(-4, -7)$ : $m_1 = \dfrac{-7 + 5}{-4 - 3} = \dfrac27$ .
$2x - 5y = 4$ has gradient $\frac25$ , so $L_2$ has $m_2 = -\frac52$ .
$\tan\theta = \left|\dfrac{m_1 - m_2}{1 + m_1m_2}\right|$
$= \left|\dfrac{\frac27 + \frac52}{1 - \frac57}\right|$
$= \dfrac{39/14}{2/7}$
$= \dfrac{39}{4}$ .
$\theta = \tan^{-1} 9.75$
$\approx 84.14^\circ$ .