3.3 - Gradients and equations of straight lines

Gradient from a graph

One ramp rises $3$ units over a horizontal run of $6$; another rises $6$ over a run of $12$. Is the second ramp steeper? Comparing the rises alone is not enough. We need the rise for each unit across.

A line's gradient is its vertical change per unit horizontal change. A rising line has a positive gradient when read from left to right; a falling line has a negative one. A horizontal line has gradient zero. A vertical line has no finite gradient because its horizontal change is zero.

Choose two well-separated, clear grid intersections on the line and make a right-angled step between them. Use coordinate units, not just numbers of squares. On one straight line, larger steps multiply both changes by the same factor, so their quotient stays the same.

[DIAGRAM: asset_slug: m048_gradient_triangle; caption: A falling straight line with a horizontal change of six units and vertical change of minus three; description: Exact line y equals minus one half x plus 2 with points A minus 2 comma 3 and B 4 comma 0, a horizontal run of 6 and a downward change of minus 3; file: m048_gradient_triangle.png; recommended_method: matplotlib]

The horizontal step in the diagram goes right, so it is positive. The vertical step goes down, so it is negative.

Change only the direction of the vertical movement next. The step has the same dimensions, but an upward step gives a positive gradient.

That check answers the ramp puzzle: doubling both dimensions keeps the same steepness. It also explains why your choice of a larger gradient triangle does not change the result.

Keep the sign as you try another falling line.

Gradient and y-intercept in an equation

The equation tells you the gradient without drawing the line. In $y=3x-4$, increasing $x$ by $1$ increases $y$ by $3$. The constant $-4$ fixes where the line meets the vertical axis.

At the $y$-axis, $x=0$, so $mx$ vanishes and $y=c$. A horizontal line $y=c$ has $m=0$. A vertical line $x=k$ cannot be written in this form with a finite gradient.

The next equation is presented differently: $y$ is not yet on its own. Rearrange first, because the coefficients in the original layout do not directly give the gradient.

You can also build an equation from its gradient and intercept: gradient $6$ through $(0,2)$ gives $y=6x+2$. Reversing the order of its terms, $y=2+6x$, changes no points on the line.

When coordinates of an intercept are requested, write a point such as $(0,5)$. The single number $5$ only gives its height.