Points, Straight Lines and Shapes in 2D

Lessons

Senior 5 · Mathematics

Points, Straight Lines and Shapes in 2D

4 lessons

Lesson 2 of 4

Intersecting, parallel and perpendicular lines; the angle between two lines

25 min

Do two roads cross?

On a town plan, two straight roads are given by their equations. Do they cross, and if so where, and at what angle? Two lines in a plane can meet in exactly one point, never meet, or be the same line. The coefficients of their equations decide which, without any drawing. This lesson gives the tests, finds intersection points, and measures the angle between two lines with the scalar product of S4 and the tangent formula of Unit 1.

Definition

Parallel lines: lines with parallel (collinear) direction vectors. Two parallel lines are either distinct (no common point) or coincident (the same line).

Perpendicular lines: lines whose direction vectors are orthogonal, $\vec{u}_1\cdot\vec{u}_2 = 0$ .

Angle between two lines: the acute (or right) angle $\theta$ , $0^\circ \le \theta \le 90^\circ$ , between them where they meet; parallel lines make the angle $0^\circ$ .

Example: $y = 2 x + 1$ and $y = 2 x - 5$ are distinct parallel lines; $y = 2 x$ and $y = -\frac{1}{2}x$ are perpendicular.

Important

Relative position of $d_{1} : a_{1} x + b_{1} y + c_{1} = 0$ and $d_{2} : a_{2} x + b_{2} y + c_{2} = 0$

Condition	Position
$a_1b_2 - a_2b_1 \ne 0$	intersecting in exactly one point
$a_{1} b_{2} - a_{2} b_{1} = 0$ , but $(a_{1}, b_{1}, c_{1})$ not proportional to $(a_{2}, b_{2}, c_{2})$	parallel and distinct
$(a_{1}, b_{1}, c_{1})$ proportional to $(a_{2}, b_{2}, c_{2})$	coincident
$a_{1} a_{2} + b_{1} b_{2} = 0$	perpendicular

Why: $a_{1} b_{2} - a_{2} b_{1}$ is the determinant of the system of the two equations (S4); it is zero exactly when the normals $(a_{1}, b_{1})$ and $(a_{2}, b_{2})$ are parallel. The last line is $\vec{n}_1\cdot\vec{n}_2 = 0$ : perpendicular normals mean perpendicular lines. When both gradients exist, these become $m_{1} = m_{2}$ (parallel) and $m_{1} m_{2} = - 1$ (perpendicular).

The angle between two lines

With direction vectors $\vec{u}_1$ and $\vec{u}_2$ , the scalar product gives the angle between the vectors; the absolute value makes it acute:

\cos\theta = \frac{|\vec{u}_1\cdot\vec{u}_2|}{\|\vec{u}_1\|\,\|\vec{u}_2\|}.

If the lines have gradients $m_1 = \tan\alpha_1$ and $m_2 = \tan\alpha_2$ , the angle between them is $|\alpha_1 - \alpha_2|$ (or its supplement), and the subtraction formula for the tangent (Unit 1) gives

\tan\theta = \left|\frac{m_1 - m_2}{1 + m_1m_2}\right| \qquad (1 + m_1m_2 \ne 0).

Figure 1. The lines of Worked example 1. The angle between two lines is the acute angle

\theta

at their intersection; here

\theta \approx 81.9^\circ

. (The obtuse angle,

180^\circ - \theta

, is the angle between

\vec{u}_1

and

\vec{u}_2

as drawn.)

Worked example 1: meet, where and at what angle

Show that $d_{1} : 2 x - y - 1 = 0$ and $d_{2} : x + 3 y - 11 = 0$ intersect. Find the point and the angle between them.

Solution $a_1b_2 - a_2b_1 = 2(3) - 1(-1) = 7 \ne 0$ : they intersect. From $d_{1}$ , $y = 2 x - 1$ ; then $x + 3 (2 x - 1) - 11 = 0$ , $7 x = 14$ , so $x = 2$ , $y = 3$ : the point is $(2, 3)$ .

Angle by gradients: $m_{1} = 2$ , $m_2 = -\frac{1}{3}$ : $\tan\theta = \left|\frac{2 + \frac{1}{3}}{1 - \frac{2}{3}}\right| = \frac{\frac{7}{3}}{\frac{1}{3}} = 7$ , $\theta = 81.9^\circ$ .

Check by vectors: $\vec{u}_1 = (1, 2)$ , $\vec{u}_2 = (-3, 1)$ : $\cos\theta = \frac{|-3 + 2|}{\sqrt{5}\sqrt{10}} = \frac{1}{\sqrt{50}} \approx 0.1414$ , $\theta = 81.9^\circ$ . ✓

Worked example 2: a parametric line meets a Cartesian line

Where does the line $x = 1 + 2 t$ , $y = - 1 + t$ meet $3 x - 4 y - 12 = 0$ ?

Solution Substitute the parametric equations: $3 (1 + 2 t) - 4 (- 1 + t) - 12 = 0$ , so $2 t - 5 = 0$ and $t = 2.5$ . The point is $(1 + 5, - 1 + 2.5) = (6, 1.5)$ . Check: $18 - 6 - 12 = 0$ . ✓

Worked example 3: parallel and perpendicular through a point

Find the lines through $P (- 1, 4)$ that are (a) parallel (b) perpendicular to $3 x + 2 y - 6 = 0$ .

Solution

Same normal $(3, 2)$ : $3 x + 2 y + c = 0$ , and $P$ gives $- 3 + 8 + c = 0$ , $c = - 5$ : $3 x + 2 y - 5 = 0$ .
The old direction $(- 2, 3)$ becomes the new normal, so use $(2, - 3)$ : $2 x - 3 y + c = 0$ , with $- 2 - 12 + c = 0$ , $c = 14$ : $2 x - 3 y + 14 = 0$ . Check: $3 (2) + 2 (- 3) = 0$ . ✓

Worked example 4: a line with a parameter

For which values of $k$ are $k x + 2 y - 3 = 0$ and $2 x + (k - 3) y + 1 = 0$ parallel? For which are they perpendicular?

Solution Parallel: $k(k - 3) - 2 \times 2 = 0$ , so $k^{2} - 3 k - 4 = 0$ and $k = 4$ or $k = - 1$ . In both cases the constants are not in the same ratio (for $k = 4$ : $\frac{4}{2} = \frac{2}{1}$ but $\frac{-3}{1} \ne 2$ ), so the lines are distinct. Perpendicular: $2 k + 2 (k - 3) = 0$ , so $k = \frac{3}{2}$ .

Warning

Gradients have limits. The formula $\tan\theta = \left|\frac{m_1 - m_2}{1 + m_1m_2}\right|$ fails for a vertical line (no gradient) and when $1 + m_{1} m_{2} = 0$ (the lines are perpendicular). Direction vectors always work. And the angle between two lines is never obtuse: if the scalar product gives $120^\circ$ between the vectors, the lines make $60^\circ$ .

Tip

Parallel or perpendicular through a point, in one line: to be parallel to $a x + b y + c = 0$ , keep $a$ and $b$ and choose a new constant; to be perpendicular, use $b x - a y + k = 0$ . Then put in the point to find the constant.

Khan Academy India: Angle between two lines. Watch how the formula for the acute angle comes from the angles the lines make with the x-axis and the formula for tan(A − B).

✦

Practice

The last two are harder.

State whether the lines intersect, are parallel and distinct, or coincide: (a) $2 x - 3 y + 1 = 0$ and $4 x - 6 y + 5 = 0$ (b) $x + 2 y - 4 = 0$ and $3 x + 6 y - 12 = 0$ (c) $x - y + 2 = 0$ and $2 x + y - 5 = 0$ .
Find the point of intersection of $3 x + 2 y = 7$ and $x - y = 4$ .
Where does the line $x = 2 - t$ , $y = 1 + 2 t$ meet $x + y - 6 = 0$ ?
Find the lines through $(2, - 3)$ parallel and perpendicular to $5 x - y + 4 = 0$ .
Find the acute angle between $y = 3 x - 1$ and $y = - x + 4$ .
Find the angle between the lines $x = 1 + t$ , $y = 2 + 2 t$ and $x = 3 - s$ , $y = 3 s$ .
Find $k$ so that $k x - 3 y + 2 = 0$ and $2 x + 4 y - 1 = 0$ are (a) perpendicular (b) parallel.
Show that the three lines $x + 2 y - 4 = 0$ , $2 x - y - 3 = 0$ and $3 x + y - 7 = 0$ pass through one point.

Show answersHide answers

(a) $2 (- 6) - 4 (- 3) = 0$ and $\frac{1}{5} \ne \frac{2}{4}$ : parallel and distinct (b) the second is 3 times the first: coincident (c) $1 \times 1 - 2 \times (-1) = 3 \ne 0$ : they intersect, at $(1, 3)$ .
$x = 3$ , $y = - 1$ .
$2 - t + 1 + 2 t - 6 = 0$ , $t = 3$ : the point $(- 1, 7)$ .
Parallel: $5 x - y - 13 = 0$ . Perpendicular: $x + 5 y + 13 = 0$ .
$\tan\theta = \left|\frac{3 - (-1)}{1 + 3(-1)}\right| = 2$ , $\theta = 63.4^\circ$ .
$\vec{u}_1 = (1, 2)$ , $\vec{u}_2 = (-1, 3)$ : $\cos\theta = \frac{5}{\sqrt{5}\sqrt{10}} = \frac{1}{\sqrt{2}}$ , $\theta = 45^\circ$ .
(a) $2 k - 12 = 0$ , $k = 6$ (b) $4 k - (- 3) (2) = 0$ , $k = -\frac{3}{2}$ .
The first two meet where $x = 4 - 2 y$ and $8 - 4 y - y - 3 = 0$ : $y = 1$ , $x = 2$ . The third: $6 + 1 - 7 = 0$ , so it passes through $(2, 1)$ too.

Sources and credits

Written by Ganzaa: the explanations, the diagram, the worked examples, the practice items and the quiz questions.
Video: Angle between two lines, Khan Academy India, embedded from YouTube.

1Do two roads cross?

2The angle between two lines

Do two roads cross?

The angle between two lines