The straight line y = mx + c

What you are looking at

The blue line is $y = mx + c$. A dashed triangle on it shows the gradient as rise over run — drag its corner anywhere along the line and the two legs change but their ratio does not, which is what “the gradient is the same everywhere on a straight line” means. The yellow dot is the $y$-intercept $(0, c)$, and the working restates the equation from those two numbers.

The Parallel and Perpendicular tabs add a point $Q$ that you can drag on the grid. The parallel line keeps the gradient $m$ and finds its own intercept from $c = y_1 - mx_1$; the perpendicular line uses $m_2 = -\dfrac{1}{m_1}$, and the working checks that $m_1 \times m_2 = -1$. Both equations are written out in full, so the method is on screen next to the picture it describes.

Try this

1. 1Slide $m$ from $\dfrac{1}{2}$ down through $0$ to $-2$: the line flattens, then tips the other way, and the rise in the gradient triangle turns negative.
2. 2On Parallel, drag $Q$ onto the blue line itself. The parallel line becomes the same line, drawn dashed, and the working says why.
3. 3On Perpendicular, set $m = 0$: the perpendicular to a horizontal line is vertical, $x =$ the $x$-coordinate of $Q$, and it has no gradient to write.
4. 4Press Play: $m$ sweeps from $-4$ to $4$ and the perpendicular line swings the opposite way, always at right angles.

Mistakes students make

• Reading the gradient off the picture when the axes have different scales. Count the squares as rise and run in units, not in centimetres.
• Confusing the $y$-intercept with the $x$-intercept: $c$ is the value of $y$ when $x = 0$. The line $y = 2x - 6$ crosses the $y$-axis at $-6$ and the $x$-axis at $3$.
• Writing the perpendicular gradient as $\dfrac{1}{m}$ or $-m$. It is $-\dfrac{1}{m}$: for $m = \dfrac{1}{2}$ the perpendicular has gradient $-2$, and the product is $-1$.

In the exam

• “Find the equation of the line through $(2, -1)$ parallel to $y = \dfrac{1}{2}x + 1$”: keep $m = \dfrac{1}{2}$, substitute the point to get $c = -1 - \dfrac{1}{2} \times 2 = -2$, and write $y = \dfrac{1}{2}x - 2$.
• Given two points, find $m = \dfrac{y_2 - y_1}{x_2 - x_1}$ first, then substitute one point to find $c$. Perpendicular lines are Extended only; parallel lines are on both papers.

Where it is taught

• Cambridge IGCSE Mathematics 0580/0980 · 3.1 Coordinates and linear graphs · 3.2 Gradient of a line · 3.4 Equation of a straight line (parallel lines; perpendicular lines: Extended)