Intercepts and Sketching
A straight line is determined by any two of its points, and two of them are usually free: the places where it crosses the axes. Finding those is one substitution each, in whichever form the equation happens to arrive.
A point on the y-axis has an x-coordinate of what? A point on the x-axis has a y-coordinate of what? Now look at any linear equation and decide what to substitute to find each crossing.
On the y-axis, $x$ is zero. On the x-axis, $y$ is zero. So substitute zero for the other variable and solve. Two crossings give two points, and two points determine the line.
$$x = 0 \to \text{y-intercept}, \qquad y = 0 \to \text{x-intercept}$$
The method works in any form, and that is its virtue. General form, gradient–intercept form, or something untidier still, substituting zero costs the same either way, so there is no need to rearrange first.
Know
- that the y-intercept is found by substituting $x = 0$, and the x-intercept by substituting $y = 0$
- that two distinct points are enough to determine and draw a straight line
- that a line through the origin, a horizontal line and a vertical line each need a different sketching approach
Understand
- why the substitution method works in any form of the equation, without rearranging first
- why the two-intercept method fails for a line through the origin
- why plotting a third point is a worthwhile check rather than wasted effort
Can Do
- find both intercepts of a line given in general form or gradient–intercept form
- sketch a line from its two intercepts, labelling both
- sketch a line from its gradient and y-intercept, and handle lines through the origin and lines parallel to an axis
An intercept is a point, not a number, although it is usually reported by the one coordinate that carries information.
The y-axis is the set of all points whose x-coordinate is zero. So a line crosses it wherever the line's own points have $x = 0$, and finding that crossing means substituting $x = 0$ into the equation and solving for $y$.
The x-axis is the set of all points whose y-coordinate is zero, so the mirror-image method applies: substitute $y = 0$ and solve for $x$.
Substitute the variable you are not looking for. Saying it that way round makes the common mix-up harder to commit.
This is the reason intercepts are worth reaching for first when a line arrives in general form.
Take $3x - 4y - 12 = 0$. Setting $y = 0$ gives $3x - 12 = 0$, so $x = 4$ and the x-intercept is $(4, 0)$. Setting $x = 0$ gives $-4y - 12 = 0$, so $y = -3$ and the y-intercept is $(0, -3)$.
Neither substitution required the equation to be rearranged. Both left a one-step linear equation.
Compare that with converting to $y = \tfrac{3}{4}x - 3$ first: correct, but three lines of algebra to reach a fact you already had. Rearrange when you need the gradient; substitute when you need the crossings.
Plot the two intercepts, rule a line through them, extend it past both, and label the line with its equation. That is the whole method.
Label the intercepts on the diagram too. A sketch showing an unlabelled line crossing the axes somewhere communicates almost nothing, whereas one marked $(4, 0)$ and $(0, -3)$ pins the line down completely.
It is worth a third point as a check when the arithmetic was fiddly. Choose any convenient $x$, work out $y$, and confirm it falls on your ruled line. Two points always give a line, but they give the wrong line without complaint if one of them is wrong. A third point has nothing to agree with unless all three are right.
Three cases defeat the method, and each has its own fix.
A line through the origin, such as $y = \tfrac{3}{4}x$. Both intercepts are the same point, $(0, 0)$, so you have one point rather than two. Fix: choose any other $x$ value and find its $y$. Choosing $x = 4$ gives $y = 3$, so use $(0, 0)$ and $(4, 3)$.
A horizontal line, such as $y = 5$. It never crosses the x-axis at all, so there is no x-intercept to find. Fix: no calculation is needed, since the line is simply the set of points five units above the axis.
A vertical line, such as $x = -2$. It has no y-intercept, for the mirror-image reason. Fix: as above, draw it directly.
Recognising these before starting saves the confusion of a method returning $0 = 0$ or an impossible statement. The equation tells you which case you are in before any working begins.
When the line is already in the form $y = mx + c$, there is a second method that avoids finding the x-intercept at all.
Plot $(0, c)$. Then use the gradient as a set of instructions: from that point, move one unit right and $m$ units up. For $y = -2x + 6$, plot $(0, 6)$, then go one right and two down to $(1, 4)$. Rule the line through the two.
For a fractional gradient, use the denominator as the run. For $m = \tfrac{3}{4}$, go four right and three up, which lands on a lattice point instead of guessing three-quarters of a square.
Both methods are correct and neither is always faster. Use intercepts when the equation is in general form, and gradient stepping when it is in gradient–intercept form. Whichever you choose, one extra point is a cheap insurance policy.
Watch Me Solve It · 3 examples
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1Find the x-intercept by setting y to zero$3x - 4(0) - 12 = 0$$3x = 12 \ \Rightarrow\ x = 4$The point is $(4, 0)$. Substituting zero removes the $y$ term entirely.
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2Find the y-intercept by setting x to zero$3(0) - 4y - 12 = 0$$-4y = 12 \ \Rightarrow\ y = -3$The point is $(0, -3)$. Divide by the negative carefully.
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3Plot, rule and labelMark $(4, 0)$ and $(0, -3)$, rule a line through them extending past both, and label it $3x - 4y - 12 = 0$.
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4Check with a third point$x = 8: \quad 24 - 4y - 12 = 0 \ \Rightarrow\ y = 3$The point $(8, 3)$ should fall on the ruled line, and it does, so the sketch is right.
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1Start at the y-interceptThe equation is in gradient–intercept form, so $c = 6$ and the line passes through $(0, 6)$ without any calculation.
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2Step by the gradient$m = -2 = \frac{-2}{1}$From $(0, 6)$, move one unit right and two units down, reaching $(1, 4)$. Rule the line through the two points.
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3Find the x-intercept to label it$0 = -2x + 6$$2x = 6 \ \Rightarrow\ x = 3$The line crosses the x-axis at $(3, 0)$. Marking it makes the sketch complete.
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1Check the intercepts before starting$x = 0 \ \Rightarrow\ y = 0$$y = 0 \ \Rightarrow\ x = 0$Both substitutions return the origin. There is only one intercept, so the two-intercept method cannot proceed.
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2Choose a second point using the denominator$x = 4: \quad y = \tfrac{3}{4}(4) = 3$Taking $x$ equal to the denominator of the gradient gives whole-number coordinates, here $(4, 3)$.
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3Rule and labelDraw the line through $(0, 0)$ and $(4, 3)$, extending it into the third quadrant as well, since the line continues in both directions.
Brain Trainer · 5 problems
Five intercept items. Give each answer as a point, and notice which of them are special cases.
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1 Find the x-intercept of $2x + 5y - 10 = 0$.
Set $y = 0$, giving $2x = 10$.$(5, 0)$ -
2 Find the y-intercept of $2x + 5y - 10 = 0$.
Set $x = 0$, giving $5y = 10$.$(0, 2)$ -
3 Find both intercepts of $y = 4x - 8$.
The y-intercept is read off; for the other, set $y = 0$.$(2, 0)$ and $(0, -8)$ -
4 Find the intercepts of $y = 7$.
A horizontal line seven units up. It never reaches the x-axis.$(0, 7)$ only; no x-intercept -
5 Find a second point to help sketch $y = -\tfrac{2}{5}x$.
It passes through the origin, so use $x = 5$ to keep the coordinates whole.$(5, -2)$
Multiple Choice · 5 questions
To find the x-intercept of a line you should:
The x-intercept of $5x + 2y - 20 = 0$ is:
A student tries to sketch $y = 2x$ using two intercepts and finds that both come out as $(0, 0)$. The best next step is to:
Sketching $y = \tfrac{2}{3}x - 1$ by stepping, from the y-intercept you should move:
A line is drawn through $(0, -4)$ and $(6, 0)$. Its equation in gradient–intercept form is:
Short Answer · 3 questions
Substitute the variable you are not looking for
Set $y = 0$ for the x-intercept and $x = 0$ for the y-intercept. Report each as a point rather than a bare number.
No rearranging needed
The substitution works in general form as easily as in gradient–intercept form, which is why it is the natural first move when a line arrives as $ax + by + c = 0$.
Three cases need different handling
A line through the origin has both intercepts at one point and needs a second point chosen freely. Horizontal and vertical lines have only one intercept each and can be drawn directly.
A third point is cheap insurance
Two points always determine a line, including the wrong line if one of them is wrong. A third point agrees only when all three are right.
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