IB Applications and Interpretation SL · 2D geometry · Section 3 of 5

Parallel and perpendicular lines

3 worked examples · By Álvaro López

Key idea

Two lines in the plane are related by their inclination: if they have the same gradient they are parallel, and if the product of their gradients is $- 1$ they are perpendicular. In the IB exam this appears when you are asked for “the line through this point that is parallel (or perpendicular) to this other one”, a step that repeats in problems about streets, fences, pipes and the perpendicular bisectors of Voronoi diagrams.

Parallel lines

Two parallel lines never meet: they have the same inclination. Since the gradient measures the inclination, it is enough to compare gradients.

Look at the figure: both lines have gradient $\frac{2}{3}$ . Every time you move 3 units to the right, both rise by 2.

[FIGURA mai-t4-a3-paralelas: Cartesian plane with a grid and labelled x and y axes, x from -4 to 8 and y from -4 to 6. Two blue parallel lines: r1 with equation y = (2/3)x - 1, which cuts the y-axis at (0,-1) and the x-axis at (1.5, 0), and r2 with equation y = (2/3)x + 2, which cuts the y-axis at (0,2). Labels "r1: y = 2x/3 - 1" and "r2: y = 2x/3 + 2" beside each line. On each line, a small stepped triangle: a horizontal run of 3 units (label "3") and a vertical rise of 2 units (label "2"), to show that both have the same gradient 2/3.]

If, in addition to the same gradient, they have the same y-intercept, they are not two different lines but the same line (coincident lines). If the lines are given in general form, $a x + b y + d = 0$ , the gradient is $m = - \frac{a}{b}$ : two lines with the same coefficients $a$ and $b$ (apart from a common factor) and a different $d$ are parallel.

Perpendicular lines

Two lines are perpendicular ( $r_{1} ⊥ r_{2}$ ) when they meet forming a right angle of $9 0^{\circ}$ .

Why does it work? A line through the origin with gradient $m$ also passes through the point $(1, m)$ . If you rotate that point through $9 0^{\circ}$ about the origin, it lands at $(- m, 1)$ , and the line through the origin and this point has gradient $\frac{1}{- m} = - \frac{1}{m}$ . That is the perpendicular.

[FIGURA mai-t4-a3-perpendiculares: Cartesian plane with a grid and labelled axes, x from -4 to 4 and y from -4 to 4. Blue line L1 through the origin with gradient 2/3, that is y = 2x/3, drawn up to (3,2); with the right-angled triangle of legs 3 (horizontal, label "3") and 2 (vertical, label "2") beneath it. Red line L2 through the origin and (-2,3), that is y = -3x/2, with its right-angled triangle of legs 2 (horizontal, to the left, label "2") and 3 (vertical, label "3"). A small right-angle square at the origin between the two lines, labelled "90°". Labels "L1: m1 = 2/3" and "L2: m2 = -3/2", and below them, in a box, "m1 · m2 = (2/3)·(-3/2) = -1".]

Notice how the triangles in the figure swap their legs (3 and 2 become 2 and 3) and one of the runs changes direction: that is where the reciprocal and the negative sign come from.

In general form it is even quicker: the line $a x + b y + d = 0$ has gradient $- \frac{a}{b}$ , and the line $b x - a y + d^{'} = 0$ has gradient $\frac{b}{a}$ . They are perpendicular: you swap $a$ and $b$ and change one sign.

Line through a point, parallel or perpendicular to another line

This is a typical exam task. It is solved with the point-gradient form, $y - y_{0} = m (x - x_{0})$ , where $(x_{0}, y_{0})$ is the given point.

Worked examples

Example 1 Basic

Streets in a neighbourhood

On the plan of a neighbourhood (one unit equals 100 m) there are three straight streets with these equations:

$r : y = - \frac{2}{5} x + 1 s : 2 x + 5 y - 10 = 0 t : 5 x - 2 y + 3 = 0$

a) Find the gradient of each street.

b) State which streets are parallel and which pairs are perpendicular.

Solution

a) Street $r$ is already in explicit form: $m_{r} = - \frac{2}{5}$ .

For $s$ , we rearrange for $y$ : $5 y = - 2 x + 10$ , so $y = - \frac{2}{5} x + 2$ and $m_{s} = - \frac{2}{5}$ .

For $t$ : $2 y = 5 x + 3$ , so $y = \frac{5}{2} x + \frac{3}{2}$ and $m_{t} = \frac{5}{2}$ .

b) $r$ and $s$ have the same gradient, $- \frac{2}{5}$ , and different y-intercepts ( $1$ and $2$ ): they are parallel.

Check the product with $t$ : $- \frac{2}{5} \cdot \frac{5}{2} = - 1$ . Therefore $t ⊥ r$ and, since $s$ has the same gradient as $r$ , also $t ⊥ s$ .

The perpendicular pairs are $r$ - $t$ and $s$ - $t$ .

Example 2 Medium

A pipe to run alongside

On a map with units of 1 km, a water pipe follows the line $L : 3 x + 4 y - 12 = 0$ . From the reservoir at $P (5, - 2)$ , a channel parallel to the pipe and an access track perpendicular to it are to be built.

a) Find the equation of the channel (line through $P$ parallel to $L$ ) in the form $a x + b y + d = 0$ with $a$ , $b$ and $d$ integers and $a > 0$ .

b) Find the equation of the track (line through $P$ perpendicular to $L$ ) in the same form.

Solution

The gradient of $L$ is $m_{L} = - \frac{3}{4}$ .

a) Parallel: same gradient, $m = - \frac{3}{4}$ . With the point $P (5, - 2)$ : $y - (- 2) = - \frac{3}{4} (x - 5) \Rightarrow 4 y + 8 = - 3 x + 15 \Rightarrow 3 x + 4 y - 7 = 0$ Check: $3 \cdot 5 + 4 \cdot (- 2) = 15 - 8 = 7$ , so $P$ satisfies the equation.

b) Perpendicular: $m = - \frac{1}{- 3/4} = \frac{4}{3}$ . With the point $P$ : $y + 2 = \frac{4}{3} (x - 5) \Rightarrow 3 y + 6 = 4 x - 20 \Rightarrow 4 x - 3 y - 26 = 0$ Check: $4 \cdot 5 - 3 \cdot (- 2) = 20 + 6 = 26$ . Also $(- \frac{3}{4}) \cdot \frac{4}{3} = - 1$ .

Example 3 Hard

Paths in a triangular park

A park is triangular in shape, with vertices $A (- 2, 1)$ , $B (4, 3)$ and $C (0, 7)$ on a plan with units of 100 m. Two straight paths are to be built: one from $C$ perpendicular to the side $A B$ , and another from $B$ perpendicular to the side $A C$ .

a) Find the equation of the path starting at $C$ in the form $a x + b y + d = 0$ , with $a$ , $b$ , $d$ integers and $a > 0$ .

b) Find the equation of the path starting at $B$ in the same form.

c) The paths cross at the point $H$ . Find its coordinates.

d) Calculate the product of the gradients of $H A$ and $B C$ and interpret the result.

[FIGURA mai-t4-a3-parque: Cartesian plane with a grid and labelled x and y axes (x from -3 to 6, y from -1 to 8). Triangle ABC with vertices marked as dots and labelled: A(-2, 1), B(4, 3) and C(0, 7). Sides drawn in grey. From C, a blue line goes down to meet side AB; from B, a red line goes towards side AC. Both lines cross at the point H(1, 4), labelled inside the triangle. Small right-angle squares where each line meets the side it is perpendicular to. Dotted segment from A to H.]

Solution

a) The gradient of $A B$ is $m_{A B} = \frac{3 - 1}{4 - ( - 2 )} = \frac{2}{6} = \frac{1}{3}$ . The path from $C$ is perpendicular to $A B$ , so its gradient is $- 3$ . Through $C (0, 7)$ : $y - 7 = - 3 (x - 0) \Rightarrow y = - 3 x + 7 \Rightarrow 3 x + y - 7 = 0$

b) The gradient of $A C$ is $m_{A C} = \frac{7 - 1}{0 - ( - 2 )} = \frac{6}{2} = 3$ . The path from $B$ is perpendicular to $A C$ : gradient $- \frac{1}{3}$ . Through $B (4, 3)$ : $y - 3 = - \frac{1}{3} (x - 4) \Rightarrow 3 y - 9 = - x + 4 \Rightarrow x + 3 y - 13 = 0$

c) The point $H$ satisfies both equations. From the first, $y = 7 - 3 x$ . Substituting into the second: $x + 3 (7 - 3 x) - 13 = 0 \Rightarrow x + 21 - 9 x - 13 = 0 \Rightarrow - 8 x + 8 = 0 \Rightarrow x = 1$ So $y = 7 - 3 = 4$ , hence $H = (1, 4)$ . (Using the GDC, it is enough to draw both lines and ask for the intersection; check that it agrees.)

d) The gradient of $H A$ is $\frac{4 - 1}{1 - ( - 2 )} = 1$ and that of $B C$ is $\frac{7 - 3}{0 - 4} = - 1$ . The product is $1 \cdot (- 1) = - 1$ , so $H A ⊥ B C$ : the path joining $A$ with $H$ would be perpendicular to the side $B C$ . The three paths from each vertex, perpendicular to the opposite side, meet at a single point, $H$ .

In the exam: each part earns marks for the calculation of the gradient, the use of the relation $m_{1} m_{2} = - 1$ and the final equation in the required form. In part c), write down the system you solve, even if you do it on the GDC.

Common mistakes

Summary

Two lines are parallel if they have the same gradient, $m_{1} = m_{2}$ (and different y-intercepts; otherwise they coincide).
Two lines are perpendicular if $m_{1} \cdot m_{2} = - 1$ , that is, $m_{2} = - \frac{1}{m _{1}}$ : invert the gradient and change its sign.
In general form, $a x + b y + d = 0$ has gradient $- \frac{a}{b}$ ; the lines $a x + b y + d = 0$ and $b x - a y + d^{'} = 0$ are perpendicular.
Vertical lines ( $x = k$ ) and horizontal lines ( $y = c$ ) are treated separately: two verticals are parallel, and a vertical line is perpendicular to a horizontal one.
To find the line through $(x_{0}, y_{0})$ parallel or perpendicular to another line: gradient $m_{1}$ of the given line, gradient $m$ of the required line, and $y - y_{0} = m (x - x_{0})$ .
Always check that the point satisfies the equation and that the product or equality of gradients is correct; on the GDC, draw both lines with the axes at the same scale.

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