Parallel and Perpendicular Lines

A line has gradient $\tfrac{2}{3}$. Sketch it, then sketch a line at right angles to it. Read the rise and run of your second line. What has happened to the fraction $\tfrac{2}{3}$?

Parallel lines have equal gradients: $m_1 = m_2$. Perpendicular lines have gradients whose product is $-1$: $m_1m_2 = -1$, so each is the negative reciprocal of the other.

Gradient measures direction, and parallel lines point the same way, so their gradients are equal. The converse holds too: two lines with equal gradients point the same way and so never meet.

One caution belongs with that statement. Equal gradients guarantee that the lines are parallel or identical. The equations $y = 2x + 1$ and $2y = 4x + 2$ have the same gradient, but they are the same line written twice, and a line is not parallel to itself.

So when equal gradients turn up, check that the lines are genuinely different. The quickest test is the y-intercept: same gradient with different intercepts means parallel, and same gradient with the same intercept means identical.

In practice this matters most in general form, where one equation can be a multiple of the other without looking like it. Doubling every coefficient of $2x - 3y + 6 = 0$ gives $4x - 6y + 12 = 0$, which is the same line.

Take a line whose rise is $a$ and whose run is $b$, so its gradient is $\dfrac{a}{b}$. Rotate that rise-and-run triangle through a right angle.

What was the run becomes the rise, and what was the rise becomes the run, so the fraction turns over. But the rotation also sends one of the two directions backwards, so exactly one of them changes sign. The new gradient is $-\dfrac{b}{a}$.

Multiplying the two together:

$\dfrac{a}{b} \times \left( -\dfrac{b}{a} \right) = -1$

So $m_1m_2 = -1$, and each gradient is the negative reciprocal of the other.

Both operations are required. The negative reciprocal of $\tfrac{2}{3}$ is $-\tfrac{3}{2}$; taking only the negative gives $-\tfrac{2}{3}$, which is a line falling at the same steepness, and taking only the reciprocal gives $\tfrac{3}{2}$, which is steeper but still rising. Neither is at a right angle.

Three steps, and the first two are usually the work.

Extract both gradients. Rearrange each equation into gradient–intercept form, or use $m = -\dfrac{a}{b}$ from Lesson 4. Nothing can be decided before this is done.

Compare. Equal gradients means parallel, provided the lines are distinct. A product of $-1$ means perpendicular. Anything else means the lines simply cross at some other angle.

Justify. A bare verdict earns little. State the two gradients, state the relationship you tested, and state the conclusion, in that order.

"Neither" is a legitimate answer and the most common one for a randomly chosen pair. Two lines meeting at $37$ degrees are neither parallel nor perpendicular, and there is nothing further to say about them.

The other half of the topic asks for an equation rather than a verdict, and it reuses Lesson 6 unchanged.

Find the gradient of the given line. If a parallel line is wanted, keep that gradient. If a perpendicular one is wanted, take its negative reciprocal. Then substitute the new gradient and the given point into $y - y_1 = m(x - x_1)$.

The point almost never lies on the original line, and it does not need to. The original line supplies only a direction; the point supplies the position.

That division of labour is worth holding onto, because it explains why the question can give you a line and a point that seem unrelated. One of them is telling you which way to go and the other is telling you where to start.

A horizontal line and a vertical line are perpendicular. Everyone can see it, and yet the product rule cannot show it.

The horizontal line has gradient $0$. The vertical line has an undefined gradient, so there is nothing to multiply by, and $0 \times \text{undefined}$ is not a calculation at all, let alone one equal to $-1$.

So the rule should be stated with its scope: for two lines that both have gradients, they are perpendicular exactly when the product of those gradients is $-1$. The horizontal-and-vertical pair sits outside the scope and is handled by inspection.

This is the same limitation met in Lessons 4 and 6, in a third disguise. Any rule built around $m$ excludes vertical lines, because vertical lines have no $m$. Recognising the pattern is more useful than memorising three separate exceptions.

Five items. For each, decide first whether the question is about equal gradients or negative reciprocals.

1. 1 What is the gradient of any line parallel to $y = 5x - 2$?

   Show answer
   Parallel means equal gradients.$5$

2. 2 What is the gradient of any line perpendicular to $y = 5x - 2$?

   Show answer
   The negative reciprocal of $5$, which is $\tfrac{5}{1}$.$-\tfrac{1}{5}$

3. 3 What is the gradient of any line perpendicular to $y = -\tfrac{2}{3}x + 1$?

   Show answer
   Flip $\tfrac{2}{3}$ and change the sign of the whole thing.$\tfrac{3}{2}$

4. 4 Are $y = 4x + 1$ and $y = -4x + 1$ perpendicular?

   Show answer
   Their product is $4 \times (-4) = -16$.No, the product is not $-1$

5. 5 Find the line through $(0, -3)$ parallel to $2x + y - 8 = 0$.

   Show answer
   $m = -\tfrac{2}{1} = -2$, and the point is the y-intercept.$y = -2x - 3$