Equation of a line calculator: y = mx + c

1. Find the gradient: $m = \dfrac{y_2 - y_1}{x_2 - x_1}$.
2. Substitute one point into $y = mx + c$ and solve for $c$.
3. Write the equation.

Example: $(1, 2)$ and $(5, 10)$. $m = 8 \div 4 = 2$. Then $2 = 2 \times 1 + c$, so $c = 0$. The line is $y = 2x$.

Through $(2, 7)$ with gradient 3: $7 = 3 \times 2 + c$, so $c = 1$ and the line is $y = 3x + 1$.

Through $(0, 4)$ with gradient $-0.5$: the point is on the y-axis, so $c = 4$ straight away: $y = -0.5x + 4$.

A line parallel to $y = 3x + 1$ has gradient 3 too. A line perpendicular to it has gradient $-\tfrac{1}{3}$, because perpendicular gradients multiply to $-1$. To find a perpendicular line through a point, use that gradient with the point method above.

Equations may be given as $ax + by = c$, like $2x + y = 8$. Rearrange to $y = mx + c$ to read the gradient and intercept: $y = -2x + 8$, so the gradient is $-2$ and it crosses the y-axis at 8. To find where a line crosses the x-axis, set $y = 0$.

• Using $(x, y)$ the wrong way round when substituting.
• Getting the sign of $c$ wrong when moving terms.
• Writing $y = 2x + 0$: just write $y = 2x$.

Practise with our straight-line graphs questions, and find the gradient alone with the gradient calculator.

• Check your equation with the second point: it should work for both.
• If the question asks for the form $ax + by + c = 0$, rearrange at the end and use whole numbers.
• On the SAT, lines are often written $y = mx + b$: same idea.

• Is a point on the line? Substitute its x-coordinate. For $y = 3x + 1$, the point $(4, 13)$ is on the line because $3 \times 4 + 1 = 13$, but $(4, 12)$ is not.
• x-intercept: set $y = 0$. For $y = 3x + 1$, $0 = 3x + 1$ gives $x = -\tfrac{1}{3}$.
• Where two lines cross: set their equations equal and solve, or use our simultaneous equations calculator.