Slopes of perpendicular lines

Why this exists

Parallel lines are easy to spot from their equations: they have the same Slope. The other special way for two lines to sit is at a right angle, and that one is harder to see. A sketch of two lines crossing at 89^\circ looks exactly like a sketch of two lines crossing at 90^\circ, and a protractor laid on a drawing measures the drawing, not the lines.

Coordinates settle it exactly. Once each line has a slope, one multiplication decides whether the lines are perpendicular, with no measuring at all. That turns questions about square corners into arithmetic: whether a triangle given by its corners has a right angle, whether four points make a rectangle, which way the tangent to a circle runs. It is the test that lets algebra prove things about right angles.

Definition

Two lines are perpendicular when they cross at a right angle. With a symbol this is written \ell_1 \perp \ell_2.

On the coordinate plane there is a test for this in terms of slope. Two lines that are not vertical, with slopes m_1 and m_2, are perpendicular exactly when their slopes multiply to -1:

Dividing both sides by m_1 (which is not zero, since the product is not zero) gives the same fact the other way round. Each slope is the negative reciprocal of the other:

To get the negative reciprocal of a number, take its reciprocal (flip the fraction) and change its sign. The negative reciprocal of \displaystyle \frac{2}{3} is \displaystyle -\frac{3}{2}, the negative reciprocal of -4 is \displaystyle \frac{1}{4}, and the negative reciprocal of 1 is -1. Both parts are needed: \displaystyle \frac{2}{3} and \displaystyle -\frac{2}{3} have opposite signs but multiply to \displaystyle -\frac{4}{9}, and \displaystyle \frac{2}{3} and \displaystyle \frac{3}{2} are reciprocals but multiply to 1.

Two kinds of line need separate treatment. A horizontal line has slope 0, and 0 has no reciprocal; a vertical line has no slope at all. A horizontal line and a vertical line are perpendicular, since the grid lines themselves meet square, but the product test says nothing about them.

Intuition

Figure 1 draws the line through V with slope \displaystyle \frac{2}{3}, together with one of its slope triangles: run 3, rise 2, ending at P(3, 2), counting from V.

Now give the whole picture a quarter turn counterclockwise about V. Every line through V turns by 90^\circ, so the turned line is perpendicular to the first one. The slope triangle turns with it and keeps its shape, but its legs change jobs. The run of 3, which went right, now goes up: it has become a rise of 3. The rise of 2, which went up, now goes left: it has become a run of -2. The point P(3, 2) lands on P'(-2, 3), and the turned line has slope

Nothing about the numbers 2 and 3 mattered. A slope triangle with rise a and run b turns into one with rise b and run -a, so a slope of \displaystyle \frac{a}{b} turns into \displaystyle \frac{b}{-a} = -\frac{b}{a}. Rise and run trading places is the reciprocal; the run that now points left is the minus sign.

This also explains how a perpendicular pair looks. A steep line has a shallow perpendicular, and the other way round, because a large number has a small reciprocal. And one line of the pair rises while the other falls, because their slopes have opposite signs.

Results

Worked examples

Boundaries

Vertical lines have no slope. The test m_1 m_2 = -1 needs two numbers, so it cannot be used when either line is vertical. A vertical line is perpendicular to exactly the horizontal lines, and that is read straight off the equations x = a and y = b.

Slope 0 has no negative reciprocal. The expression \displaystyle -\frac{1}{0} is not a number. The perpendicular to a horizontal line is vertical, which is a direction with no slope, not a line with a very large slope.

The product must be exactly -1. Slopes of 0.33 and -3 multiply to -0.99, and lines with those slopes miss a right angle by a little. Rounding a slope can turn a perpendicular pair into one that is not, so the slopes are kept as exact fractions.

Both axes must use the same unit. Slope compares a distance up with a distance across, and a right angle is a fact about true distances. If a graph stretches one axis, perpendicular lines are drawn at some other angle, though the algebra is unchanged. When the two axes measure different things, such as hours across and kilometers up, a right angle on the graph has no meaning, and the product of the slopes has nothing to report.

The lines lie in one plane. The test is about lines in the coordinate plane. In space a line has no single slope, and two lines can miss each other without being parallel; right angles there are tested with the dot product instead.

Segments are tested through their lines. Two segments are called perpendicular when the lines containing them are, even if the segments themselves do not touch.

Where this goes next

The most direct use is writing the line through a given point perpendicular to a given line: the negative reciprocal supplies the slope, and point-slope form does the rest. That in turn gives the perpendicular bisector of a segment from its endpoints, the altitudes of a triangle and the point where they meet, its orthocenter, and the distance from a point to a line, which is measured along the perpendicular.

In coordinate proofs the slope test is how a right angle gets proved. A rectangle, a right triangle, a rhombus with perpendicular diagonals: each comes down to multiplying two slopes.

Later the same condition returns in another form. A line of slope m runs in the direction of the arrow (1, m), one step across and m up. For two such arrows,

and this is zero exactly when m_1 m_2 = -1. That sum is the dot product of the two arrows, and vectors at right angles are exactly the ones whose dot product is zero. That version needs no slopes, so it works for vertical lines too, and in space.

Exercises

Negative reciprocals

Find the slope of a line perpendicular to a line with the given slope.

1. 5
2. \displaystyle -\frac{3}{4}
3. \displaystyle \frac{1}{6}
4. -1
5. 0.4
6. 0

Perpendicular or not

Decide whether the two lines are perpendicular.

7. y = 4x + 1 and \displaystyle y = -\frac{1}{4}x - 3
8. \displaystyle y = \frac{3}{2}x and \displaystyle y = \frac{2}{3}x + 5
9. x - 2y = 6 and 2x + y = 1
10. 3x + 4y = 12 and 4x + 3y = 12
11. y = 5 and x = -2
12. The line through (-1, 1) and (2, 7), and the line through (0, 3) and (6, 0)

Right angles from coordinates

13. Triangle ABC has corners A(0, 0), B(4, 2) and C(1, 8). Find the slope of each side, and decide where, if anywhere, the triangle has a right angle.
14. Triangle PQR has corners P(-1, 3), Q(2, 1) and R(5, 4). Is it a right triangle?
15. Show that A(1, 1), B(5, 3), C(3, 7) and D(-1, 5), taken in order, are the corners of a rectangle. Is it a square?
16. Quadrilateral EFGH has corners E(0, 0), F(3, 1), G(4, 4) and H(1, 3). Find the slopes of its diagonals EG and FH. Are the diagonals perpendicular?

Unknown coefficients

17. For what value of k is y = kx + 9 perpendicular to y = 6x - 2?
18. For what value of a is ax + 5y = 2 perpendicular to y = 4x - 3?
19. The line through (2, t) and (6, 5) is perpendicular to y = -2x + 7. Find t.
20. For which values of k are y = kx + 2 and 3y = (3k - 10)x + 1 perpendicular?
21. The points O(0, 0), A(4, 2) and P(a, 6) form a triangle. For what value of a does the triangle have a right angle at A?

Reasoning

22. Explain why two lines that both have positive slopes can never be perpendicular.
23. Lines p and q are both perpendicular to a line \ell of slope m \ne 0. Show that p and q are parallel or are the same line.
24. Show that Ax + By = C and Bx - Ay = D are perpendicular for any numbers A and B that are not both zero, including when one of them is zero.
25. Slopes of 0.33 and -3 look like a perpendicular pair, since 0.33 is close to \displaystyle \frac{1}{3}. Are lines with these slopes perpendicular? Explain.