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Angle between two lines

When two non-parallel lines meet, they form four angles in two pairs of equal opposite angles. We take the acute angle as the angle between them. The formula expresses this angle via the slopes.

The angle formula

Suppose two lines have slopes m1,m2m_1, m_2 and make angles θ1,θ2\theta_1, \theta_2 with the positive xx-axis. Then m1=tanθ1,m2=tanθ2m_1 = \tan\theta_1, m_2 = \tan\theta_2. The angle between them is θ=θ2θ1\theta = \theta_2 - \theta_1 (or its supplement). Using the tangent-difference formula: tanθ=tanθ2tanθ11+tanθ2tanθ1=m2m11+m1m2.\tan\theta = \frac{\tan\theta_2 - \tan\theta_1}{1 + \tan\theta_2 \tan\theta_1} = \frac{m_2 - m_1}{1 + m_1 m_2}.

To always get the acute angle, take the absolute value: tanθ=m1m21+m1m2.\boxed{\tan\theta = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right|.}

Special cases

Parallel. Slopes equal: m1=m2m_1 = m_2. The formula gives tanθ=0θ=0\tan\theta = 0 \Rightarrow \theta = 0.

Perpendicular. tanθ=\tan\theta = \infty requires the denominator to vanish: 1+m1m2=01 + m_1 m_2 = 0, i.e. m1m2=1.\boxed{m_1 m_2 = -1.}

This fails when one line is vertical (slope undefined). For verticals, perpendicular means the other is horizontal.

In terms of general form

If lines are A1x+B1y+C1=0A_1 x + B_1 y + C_1 = 0 and A2x+B2y+C2=0A_2 x + B_2 y + C_2 = 0:

  • Parallel: A1B2=A2B1A_1 B_2 = A_2 B_1.
  • Perpendicular: A1A2+B1B2=0A_1 A_2 + B_1 B_2 = 0.
  • Angle: tanθ=A1B2A2B1A1A2+B1B2\tan\theta = \left|\dfrac{A_1 B_2 - A_2 B_1}{A_1 A_2 + B_1 B_2}\right|.

Worked examples

Example 1. Find the angle between y=2x+1y = 2x + 1 and y=3x+5y = -3x + 5.

m1=2,m2=3m_1 = 2, m_2 = -3. tanθ=2(3)1+2(3)=55=1θ=45°\tan\theta = \left|\dfrac{2 - (-3)}{1 + 2 \cdot (-3)}\right| = \dfrac{5}{5} = 1 \Rightarrow \theta = 45°.

Example 2. Find the angle between 3x4y+7=03 x - 4 y + 7 = 0 and 4x+3y5=04 x + 3 y - 5 = 0.

A1A2+B1B2=1212=0A_1 A_2 + B_1 B_2 = 12 - 12 = 0. So perpendicular: θ=90°\theta = 90°.

Example 3. Find the equation of the line through (2,3)(2, 3) perpendicular to 4x+3y+7=04 x + 3 y + 7 = 0.

The given line has slope 4/3-4/3, so the perpendicular slope is 3/43/4. Equation: y3=34(x2)3x4y+6=0y - 3 = \dfrac{3}{4}(x - 2) \Rightarrow 3 x - 4 y + 6 = 0.

Example 4. Find the equations of the lines passing through (2,3)(2, 3) and making an angle of 45°45° with y=2x+5y = 2 x + 5.

m1=2m_1 = 2. tan45°=2m21+2m2=1\tan 45° = \left|\dfrac{2 - m_2}{1 + 2 m_2}\right| = 1. Case (a): 2m21+2m2=12m2=1+2m2m2=1/3\dfrac{2 - m_2}{1 + 2 m_2} = 1 \Rightarrow 2 - m_2 = 1 + 2 m_2 \Rightarrow m_2 = 1/3. Case (b): 2m21+2m2=12m2=12m2m2=3\dfrac{2 - m_2}{1 + 2 m_2} = -1 \Rightarrow 2 - m_2 = -1 - 2 m_2 \Rightarrow m_2 = -3.

Equations: y3=(1/3)(x2)x3y+7=0y - 3 = (1/3)(x - 2) \Rightarrow x - 3y + 7 = 0; y3=3(x2)3x+y9=0y - 3 = -3(x - 2) \Rightarrow 3x + y - 9 = 0.

Example 5. If the angle between y=mx+5y = mx + 5 and y=x+1y = x + 1 is tan1(1/3)\tan^{-1}(1/3), find mm.

m11+m=1/3\left|\dfrac{m - 1}{1 + m}\right| = 1/3.

Case (a): 3(m1)=1+m2m=4m=23(m - 1) = 1 + m \Rightarrow 2 m = 4 \Rightarrow m = 2. Case (b): 3(m1)=(1+m)4m=2m=1/23(m - 1) = -(1 + m) \Rightarrow 4 m = 2 \Rightarrow m = 1/2.

Try it yourself

  1. Angle between y=xy = x and y=xy = -x.
  2. Angle between y=2x+3y = 2 x + 3 and y=5y = 5.
  3. Find slope of a line perpendicular to 4x+5y8=04 x + 5 y - 8 = 0.
  4. Are 2x+3y1=02 x + 3 y - 1 = 0 and 6x+9y5=06 x + 9 y - 5 = 0 parallel?
  5. Find kk such that 3xky+2=03 x - k y + 2 = 0 is perpendicular to 2x+5y1=02 x + 5 y - 1 = 0.
  6. Find the equation of the line through (1,2)(1, 2) at 30°30° to the xx-axis.
  7. Find the equation of the line through origin perpendicular to y=3x1y = 3 x - 1.
  8. Two lines through (2,3)(2, 3) make an angle of 60°60° with y=xy = x. Find them.
  9. Show that the lines x2y+3=0x - 2 y + 3 = 0 and 2x+y5=02 x + y - 5 = 0 are perpendicular.
  10. Find the equations of the angle bisectors of 3x4y+7=03 x - 4 y + 7 = 0 and 4x3y8=04 x - 3 y - 8 = 0. (Hint: 3x4y+75=±4x3y85\dfrac{3x - 4y + 7}{5} = \pm \dfrac{4x - 3y - 8}{5}.)
  11. Find the line through (1,1)(1, 1) which is perpendicular to the line joining (2,3)(2, 3) and (5,7)(5, 7).
  12. Find θ\theta between 3x+4y7=03 x + 4 y - 7 = 0 and 5x12y+6=05 x - 12 y + 6 = 0.

Pitfalls / Tricks

  • Always take the absolute value in tanθ\tan\theta to get the acute angle.
  • m1m2=1m_1 m_2 = -1 only works when both slopes are finite.
  • When one line is vertical, the angle equals 90°θother|90° - \theta_{\text{other}}|.
  • Insight. The formula tanθ=(m1m2)/(1+m1m2)\tan\theta = |(m_1 - m_2)/(1 + m_1 m_2)| is essentially the tangent subtraction identity in disguise , derived from θ=θ1θ2\theta = \theta_1 - \theta_2 on the unit circle.

Practice quiz

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Quick check : Angle between lines
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