Graphing Straight Lines Using Intercepts
Every question with a fully worked solution.
- Graphing Straight Lines Using Intercepts - Video - Finding intercepts from an equation Watch
Theory
A linear equation such as \(y=-2x+3\) or \(3x-4y=12\) graphs as a straight line. Plot points from a table of values, or find where the line crosses the axes: the intercepts. It covers the equations of the axes, testing points by substitution, finding a missing coordinate or coefficient, and what intercepts mean in real problems.
A linear equation in \(x\) and \(y\) has no powers, products or roots of the variables. Its graph is a straight line, and every point on the line makes the equation true.
A table of values gives points to plot. For a line, equal steps in \(x\) give equal steps in \(y\). Plot at least three points: the third is a check.
The \(x\)-intercept is where the line crosses the \(x\)-axis, so \(y=0\) there. The \(y\)-intercept is where it crosses the \(y\)-axis, so \(x=0\). Two intercepts are enough to rule the line.
The line \(x=k\) is vertical and the line \(y=c\) is horizontal. The \(x\)-axis itself is \(y=0\) and the \(y\)-axis is \(x=0\).
A point lies on a line if its coordinates satisfy the equation: substitute and check that the left-hand side equals the right-hand side.
| To find | Do this |
|---|---|
| \(x\)-intercept | put \(y=0\) and solve for \(x\) |
| \(y\)-intercept | put \(x=0\) and solve for \(y\) |
| a point is on the line | substitute; LHS must equal RHS |
| a missing coefficient | substitute a known point and solve |
Special lines:
How to graph a straight line
- Choose a method: a table of values for \(y = \ldots\), or the two intercepts for a form such as \(ax+by=c\).
- Calculate at least three points, or the two intercepts plus one check point.
- Plot the points on a number plane with labelled, evenly scaled axes.
- Rule a straight line through them, extended past the points, and label it with its equation.
Problems built on graphs of lines
- Is the point on the line? Substitute both coordinates (Example 2). One mismatch is enough to say no.
- Missing coordinate or coefficient: substitute the point and solve for the unknown (Example 3).
- Area with the axes: find both intercepts, then use half base times height with positive lengths (Example 1). If the area is given, work backwards to the unknown.
- Equation from intercepts: two intercepts give two points; find the gradient, then write the equation. Clear fractions to reach integer form \(ax+by=c\).
- Context: an intercept on the time axis is when the quantity reaches zero; the intercept on the other axis is the starting value (Example 4).
- Table with an unknown: for a straight line the \(y\)-values change in equal steps for equal steps in \(x\). Set two steps equal and solve.
(i) find the \(x\)- and \(y\)-intercepts
(ii) sketch the line
(iii) find the area of the triangle it makes with the axes.
(i) The equation is not in the form \(y=mx+c\), so substitute. Zero the other letter each time.
| \(y=0:\quad 3x\) | \(=\) | \(12\) |
| \(x\) | \(=\) | \(4\) |
| \(x=0:\quad -4y\) | \(=\) | \(12\) |
| \(y\) | \(=\) | \(-3\) |
(ii) Plot \((4,0)\) and \((0,-3)\) and rule a line through them.
(iii) The triangle has a right angle at the origin, a base of \(4\) and a height of \(3\). Lengths are positive, even though the intercept is \(-3\).
| \(\text{area}\) | \(=\) | \(\dfrac{1}{2}\times 4\times 3\) |
| \(\) | \(=\) | \(\mathbf{6}\text{ square units}\) |
(i) Complete a table of values for \(y=-2x+3\) for \(x=-1,0,1,2,3\) and graph the line.
(ii) Does the point \((4,-6)\) lie on the line?
(i) Substitute each \(x\)-value.
| \(x\) | \(-1\) | \(0\) | \(1\) | \(2\) | \(3\) |
|---|---|---|---|---|---|
| \(y\) | \(5\) | \(3\) | \(1\) | \(-1\) | \(-3\) |
Each step of \(1\) in \(x\) changes \(y\) by \(-2\): equal steps mean a straight line.
(ii) Substitute \(x=4\) and compare with \(y=-6\).
| \(\text{RHS}\) | \(=\) | \(-2\times 4+3 = -5\) |
| \(\text{LHS}\) | \(=\) | \(-6\) |
| \(-6\) | \(\ne\) | \(-5\) |
The two sides are not equal, so \((4,-6)\) is not on the line. The point on the line at \(x=4\) is \((4,-5)\).
(i) Find \(a\).
(ii) Find the intercepts of the line.
(i) The point is on the line, so its coordinates make the equation true.
| \(a(4)-2(-1)\) | \(=\) | \(10\) |
| \(4a+2\) | \(=\) | \(10\) |
| \(4a\) | \(=\) | \(8\) |
| \(a\) | \(=\) | \(\mathbf{2}\) |
(ii) The line is \(2x-2y=10\).
| \(y=0:\quad 2x\) | \(=\) | \(10,\ \ x = 5\) |
| \(x=0:\quad -2y\) | \(=\) | \(10,\ \ y = -5\) |
The \(x\)-intercept is \(5\) and the \(y\)-intercept is \(-5\).
(i) What does the \(F\)-intercept mean?
(ii) Find the \(t\)-intercept and say what it means.
(i) At \(t=0\), \(F=52\). The \(F\)-intercept is the fuel in the tank at the start: a full \(52\) L.
(ii) The \(t\)-intercept is where \(F=0\).
| \(0\) | \(=\) | \(52-6.5t\) |
| \(6.5t\) | \(=\) | \(52\) |
| \(t\) | \(=\) | \(\mathbf{8}\) |
The tank runs empty after \(8\) hours of driving.
Only the part of the line from \(t=0\) to \(t=8\) makes sense here: time cannot be negative and the tank cannot hold less than nothing.
Common pitfalls
Frequently asked questions
How do you find the x- and y-intercepts of a line?
For the \(x\)-intercept put \(y=0\) and solve for \(x\). For the \(y\)-intercept put \(x=0\) and solve for \(y\).
How many points do you need to graph a straight line?
Two points fix a straight line, but plot a third as a check. If the three points are not in line, there is a mistake.
What does the graph of x = 3 look like?
A vertical line crossing the \(x\)-axis at \(3\). Every point on it has \(x\)-coordinate \(3\), whatever its \(y\)-coordinate.
How do you check if a point is on a line?
Substitute the point's \(x\) and \(y\) into the equation. If the left side equals the right side, the point is on the line; if not, it is not.
What is the equation of the x-axis?
\(y=0\), because every point on the \(x\)-axis has a \(y\)-coordinate of \(0\). The \(y\)-axis is \(x=0\).