Find the equation of a straight line from given conditions, and use gradients to test for parallel and perpendicular lines

What this dot point is asking

SEAB wants you to find the equation of a straight line from given conditions in the coordinate plane, and to use gradients to decide whether two lines are parallel or perpendicular. This applies the straight-line ideas from the graphs strand to coordinate geometry problems.

The answer

The equation of a line

A straight line has equation $y = mx + c$, with gradient $m$ and $y$-intercept $c$. To determine it you need either two points, or one point and the gradient. From two points, compute the gradient, then substitute one point to find $c$.

Using a point and a gradient

Given a gradient $m$ and a point $(x_1, y_1)$, substitute directly into $y = mx + c$ to find $c$, or use the point-gradient relationship $y - y_1 = m(x - x_1)$ and rearrange into the required form. Both routes give the same equation.

Parallel lines

Two lines are parallel exactly when their gradients are equal. So a line parallel to $y = 3x - 1$ has gradient $3$, differing only in its intercept. Comparing gradients is the test for parallelism.

Perpendicular lines

Two lines are perpendicular when the product of their gradients is $-1$:

So the gradient of a perpendicular line is the negative reciprocal of the original gradient. A line perpendicular to one of gradient $\dfrac{2}{3}$ has gradient $-\dfrac{3}{2}$.

Finding where two lines intersect

A natural follow-on is finding the point where two lines cross, which you do by solving their equations simultaneously. Set the two expressions for $y$ equal (or use elimination), solve for $x$, then substitute back to get $y$. The intersection of $y = 2x + 1$ and $y = -x + 7$ comes from $2x + 1 = -x + 7$, so $3x = 6$, $x = 2$, and $y = 5$, giving the point $(2, 5)$. This single skill underlies many coordinate-geometry tasks, such as finding the foot of a perpendicular or the vertex of a shape, because those points are always intersections of two lines you can write down.

Reading the gradient from the general form

E-Maths sometimes gives a line as $ax + by = c$ rather than $y = mx + c$, and you cannot read the gradient off it directly. Rearrange to make $y$ the subject first: from $3x + 2y = 12$, you get $y = -\tfrac{3}{2}x + 6$, so the gradient is $-\tfrac{3}{2}$. Only after this rearrangement can you apply the parallel or perpendicular tests. Converting any line into gradient-intercept form before comparing gradients is a small but essential habit, since a sign error during rearrangement would flip every later conclusion about parallel or perpendicular.

Examples in context

Example 1. The perpendicular from a point. Finding the shortest path from a point to a line means constructing the perpendicular line through the point, which uses the negative-reciprocal gradient. This is the basis of finding a perpendicular distance.

Example 2. Sides of a rectangle. In a rectangle drawn on coordinate axes, adjacent sides are perpendicular and opposite sides are parallel. Checking gradients confirms the shape, a common coordinate-geometry verification.

Try this

Q1. State the gradient of a line perpendicular to $y = 4x - 3$. [1 mark]

• Cue. Negative reciprocal of $4$ is $-\dfrac{1}{4}$.

Q2. Find the equation of the line with gradient $-3$ through $(2, 1)$. [2 marks]

• Cue. $1 = -3(2) + c$, so $c = 7$, giving $y = -3x + 7$.

Q3. Are the lines $y = 2x + 1$ and $y = 2x - 5$ parallel, perpendicular or neither? [1 mark]

• Cue. Equal gradients of $2$, so they are parallel.