Use the equation $y = mx + c$ to find the gradient and intercept; find the equation of a line through given points; and identify parallel and perpendicular lines (perpendicular at Higher tier).

What this dot point is asking

The Eduqas algebra content asks you to work with the straight line $y = mx + c$: to read off the gradient and intercept, to find the equation of a line through given points, and to identify parallel and perpendicular lines (perpendicular at Higher tier). Straight-line graphs connect algebra to coordinate geometry and underpin distance-time and conversion graphs, so they appear on both components. The perpendicular-line question is a dependable Higher-tier task because it tests the negative-reciprocal rule alongside the equation method.

The equation y = mx + c

Every straight line (except a vertical one) can be written as $y = mx + c$.

To read the gradient and intercept, the equation must be in this form. So $2y = 6x + 8$ first becomes $y = 3x + 4$, giving gradient $3$ and intercept $4$. A vertical line has the form $x = a$ and a horizontal line $y = b$.

Finding the gradient from two points

The gradient between two points is the rise divided by the run. For $(1, 2)$ and $(4, 11)$, the gradient is $\dfrac{11 - 2}{4 - 1} = \dfrac{9}{3} = 3$. Subtract the coordinates in the same order top and bottom, or the sign will be wrong.

Finding the equation of a line

Given a gradient and a point, or two points, you can find the full equation.

Parallel and perpendicular lines

Two lines are parallel when they have the same gradient, so $y = 4x + 1$ and $y = 4x - 3$ never meet. Perpendicular lines (a Higher-tier focus) cross at a right angle.

To find a line perpendicular to $y = 3x + 2$ through $(6, 1)$, the perpendicular gradient is $-\tfrac{1}{3}$. Substituting the point, $1 = -\tfrac{1}{3}(6) + c$, so $c = 1 + 2 = 3$ and the line is $y = -\tfrac{1}{3}x + 3$.

Drawing and reading a line

To draw a line from its equation, build a small table of values: choose three convenient $x$ values, work out each $y$, plot the points and join them with a ruler. Three points (rather than two) give a built-in check, because a slip shows up as a point off the line. To find where a line crosses the axes, set $x = 0$ for the $y$-intercept and $y = 0$ for the $x$-intercept. For $y = 2x - 6$, the $y$-intercept is $(0, -6)$ and setting $y = 0$ gives $x = 3$, so the $x$-intercept is $(3, 0)$.

Why straight lines matter

The straight line is the model behind real-world relationships with a constant rate: a taxi fare with a fixed charge plus a rate per mile, a phone tariff, or a conversion between two units. In each, the gradient is the rate and the intercept is the fixed starting value, which is exactly what Eduqas tests when a context question asks you to interpret $m$ and $c$ in words. The same equation also reappears in distance-time graphs (where the gradient is speed) and as the linear half of a linear-quadratic simultaneous pair, so fluency with $y = mx + c$ pays off well beyond pure algebra.