COORDINATE GEOMETRY · HL
Angle between 2 lines
Find the angle from the two slopes.
Section 1 of 5
Two lines, one angle
When two lines cross, they make four angles — two acute and two obtuse — adding in pairs to $180°$.
Call the slopes $m_1$ and $m_2$.
Line P has slope $m_1$.
Line Q has slope $m_2$.
The formula
★$\tan\theta = \pm \dfrac{m_1 - m_2}{1 + m_1 m_2}$
The $\pm$ lets you choose which angle you want. Three quick consequences follow.
(i) Parallel lines
$m_1 = m_2$
$\tan\theta = \pm \dfrac{m_1 - m_2}{1 + m_1 m_2} = \pm \dfrac{0}{1 + m_1 m_2} = 0$
$\theta = 0°$ — the lines don't meet.
(ii) Perpendicular lines
$\theta = 90°$
$\tan 90° = \infty = \dfrac{1}{0}$
For the formula to "blow up" to infinity, the denominator must be zero:
$1 + m_1 m_2 = 0$
$m_1 m_2 = -1$
(iii) Acute or obtuse — the $\pm$ sign
$\theta + \alpha = 180°$
When the formula gives a positive answer (the $+$ sign), $\theta$ is the acute angle.
When it gives a negative answer (the $-$ sign), the angle is obtuse.
So pick the sign that matches what you've been asked for: $+$ for acute, $-$ for obtuse.
Section 2 of 5
Acute angle — first example
Find the acute angle between $2x - y = 3$ and $5x + y = 7$.
Reminder: from $ax + by + c = 0$, the slope is $m = -\dfrac{a}{b}$.
$2x - y = 3$
$m_1 = \dfrac{-2}{-1} = 2$
$5x + y = 7$
$m_2 = \dfrac{-5}{1} = -5$
$\tan\theta = \pm \dfrac{m_1 - m_2}{1 + m_1 m_2}$
$\tan\theta = \pm \dfrac{2 - (-5)}{1 + 2(-5)}$
$\tan\theta = \pm \dfrac{7}{-9}$
We're asked for the acute angle, so take $\tan\theta$ positive:
$\tan\theta = \dfrac{7}{9}$
$\theta = 38°$
Obtuse partner: $180° - 38° = 142°$.
Section 3 of 5
Obtuse angle
Find the obtuse angle between $2x - 5y = 7$ and $3x + 8y = 9$.
$2x - 5y = 7$
$m_1 = \dfrac{-2}{-5} = \dfrac{2}{5}$
$3x + 8y = 9$
$m_2 = -\dfrac{3}{8}$
$\tan\theta = \pm \dfrac{m_1 - m_2}{1 + m_1 m_2}$
$\tan\theta = \pm \dfrac{\dfrac{2}{5} - \left(-\dfrac{3}{8}\right)}{1 + \dfrac{2}{5}\left(-\dfrac{3}{8}\right)}$
Top: $\dfrac{2}{5} + \dfrac{3}{8} = \dfrac{16}{40} + \dfrac{15}{40} = \dfrac{31}{40}$
Bottom: $1 - \dfrac{6}{40} = \dfrac{34}{40}$
$\tan\theta = \pm \dfrac{31/40}{34/40} = \pm \dfrac{31}{34}$
Find the acute angle first:
$\tan\theta = \dfrac{31}{34}$
$\theta = 42°$
The question asked for the obtuse angle:
$180° - 42° = 138°$
Section 4 of 5
Lines at a given angle
Find the two lines through $(1, -3)$ which cut $6x + y - 7 = 0$ at $45°$.
$6x + y - 7 = 0$
$m_1 = \dfrac{-6}{1} = -6$
Let $m$ = slope of the new line, so $m_2 = m$.
$\theta = 45°$, so $\tan 45° = 1$
$\tan\theta = \pm \dfrac{m_1 - m_2}{1 + m_1 m_2}$
$1 = \pm \dfrac{-6 - m}{1 - 6m}$
$1 - 6m = \pm (-6 - m)$
Case 1 — the $+$ sign
$1 - 6m = -6 - m$
$7 = 5m$
$m = \dfrac{7}{5}$
Case 2 — the $-$ sign
$1 - 6m = -(-6 - m) = 6 + m$
$-5 = 7m$
$m = -\dfrac{5}{7}$
The two lines
Point-slope through $(1, -3)$: $\;y + 3 = m(x - 1)$.
$y + 3 = \dfrac{7}{5}(x - 1)$
$y + 3 = -\dfrac{5}{7}(x - 1)$
Section 5 of 5
Given sin θ — find tan θ first
Two lines cross at angle $\theta$. Given $\sin\theta = \dfrac{3}{5}$, and one line is $5x - 2y = 7$, find two possible equations for the second line through $(1, -7)$.
The formula needs $\tan\theta$, so convert first. Use a 3-4-5 right triangle.
$\sin\theta = \dfrac{3}{5}, \quad \cos\theta = \dfrac{4}{5}, \quad \tan\theta = \dfrac{3}{4}$
$5x - 2y = 7$
$m_1 = \dfrac{-5}{-2} = \dfrac{5}{2}$
Let $m_2 = m$ (unknown).
$\tan\theta = \pm \dfrac{m_1 - m_2}{1 + m_1 m_2}$
$\dfrac{3}{4} = \pm \dfrac{\dfrac{5}{2} - m}{1 + \dfrac{5}{2}m}$
Cross-multiply: $\;3\left(1 + \dfrac{5}{2}m\right) = \pm \, 4\left(\dfrac{5}{2} - m\right)$
Multiply through by 2 to clear the halves:
$3(2 + 5m) = \pm \, 4(5 - 2m)$
$6 + 15m = \pm (20 - 8m)$
Case 1 — the $+$ sign
$6 + 15m = 20 - 8m$
$23m = 14$
$m = \dfrac{14}{23}$
Case 2 — the $-$ sign
$6 + 15m = -20 + 8m$
$7m = -26$
$m = -\dfrac{26}{7}$
The two lines
Point-slope through $(1, -7)$: $\;y + 7 = m(x - 1)$.
$y + 7 = \dfrac{14}{23}(x - 1)$
$y + 7 = -\dfrac{26}{7}(x - 1)$
SUM
The lot in one box
Angle between 2 lines — toolkit
1.From $ax + by + c = 0$: $m = -\dfrac{a}{b}$.
2.$\tan\theta = \pm \dfrac{m_1 - m_2}{1 + m_1 m_2}$
3.Parallel: $m_1 = m_2 \;\Rightarrow\; \tan\theta = 0$.
4.Perpendicular: $m_1 m_2 = -1$.
5.$+$ sign gives the acute angle; $-$ sign gives the obtuse.
6.Or just compute the acute, then obtuse $= 180° - $ acute.
7.If given $\sin$ or $\cos$, convert to $\tan$ first (3-4-5, etc.).
End of lesson
Angle between 2 lines — HL · Mathslive.ie