Use the gradient conditions for parallel and perpendicular lines to find equations of lines and solve geometry problems

What this dot point is asking

SEAB wants you to use the gradient relationships for parallel and perpendicular lines: parallel lines share the same gradient, and perpendicular lines have gradients whose product is $-1$ (each is the negative reciprocal of the other). With these two facts plus a point, you can write the equation of a line, and you can solve geometry problems such as finding a perpendicular bisector or the normal to a curve.

The answer

Parallel lines

Two lines are parallel exactly when they have the same gradient:

So to find a line parallel to a given one, copy its gradient and then use a point the new line must pass through.

Perpendicular lines

Two lines are perpendicular (meet at a right angle) exactly when the product of their gradients is $-1$:

In words, the perpendicular gradient is the negative reciprocal: flip the fraction and change the sign. For example, the perpendicular to gradient $\tfrac{2}{3}$ is $-\tfrac{3}{2}$.

Reading the gradient from an equation

If a line is given as $y = mx + c$, the gradient is the number $m$ in front of $x$. If it is given as $ax + by + c = 0$, rearrange into $y = mx + c$ first to read the gradient. Always extract the gradient cleanly before applying the parallel or perpendicular condition.

Building the new equation

Once you have the required gradient and a point, substitute into the point-gradient form $y - y_1 = m(x - x_1)$ and rearrange. This is the same final step used throughout coordinate geometry.

Examples in context

Example 1. The normal to a curve. In calculus, the normal at a point is the line perpendicular to the tangent there. Once you find the tangent's gradient by differentiation, the normal's gradient is its negative reciprocal, so the perpendicular rule is exactly what links coordinate geometry to the calculus strand.

Example 2. Right angles in a shape. To prove that a triangle drawn on a coordinate grid is right-angled, show that two of its sides have gradients whose product is $-1$. The perpendicular condition turns a geometric claim about angles into a quick algebraic check.

Try this

Q1. Write down the gradient of any line parallel to $y = -4x + 7$. [1 mark]

• Cue. Same gradient: $-4$.

Q2. Find the gradient of a line perpendicular to one with gradient $\dfrac{3}{5}$. [1 mark]

• Cue. Negative reciprocal: $-\dfrac{5}{3}$.

Q3. Find the equation of the line through $(0, 1)$ perpendicular to $y = 2x + 3$. [3 marks]

• Cue. Perpendicular gradient $-\tfrac{1}{2}$; through $(0, 1)$ gives $y = -\tfrac{1}{2}x + 1$.