Two intercepts are usually enough to draw a line. Two gradients are enough to tell whether two lines never meet, or meet at a right angle.
Today's hook, Perpendicular gradients multiply to $-1$. That one fact turns questions about right angles into questions about arithmetic, and it appears in coordinate geometry, calculus and vectors alike.
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Recall, your gut answer first
Meet the destination, bring back what you already know, and gather the terms and formulas this lesson leans on.
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Recall, your gut answer first
+5 XP warm-up
Sketch $y = 2x + 4$ using only its two intercepts. Then sketch a line through the origin that looks perpendicular to it. What do you think its gradient is?
Before you work it out, what is your instinct? Write it down, then check it against the lesson.
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Intercepts to draw it, gradients to compare it
Work through the core explanation before applying it.
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Intercepts to draw it, gradients to compare it
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The $x$-intercept is where $y = 0$ and the $y$-intercept is where $x = 0$. Plot both and join them. To compare two lines, compare gradients: equal means parallel, and a product of $-1$ means perpendicular.
$x$-intercept: set $y = 0$ $y$-intercept: set $x = 0$ parallel: $m_1 = m_2$ perpendicular: $m_1 m_2 = -1$
Two points are enough
A straight line needs only two plotted points. The intercepts are usually the easiest pair to find.
Perpendicular means negative reciprocal
If $m_1 = \tfrac{3}{4}$ then $m_2 = -\tfrac{4}{3}$: flip the fraction and change the sign. Both steps, not one.
Through the origin, use the gradient
If the line passes through $(0,0)$ both intercepts are the same point, so plot the origin and step out one gradient to get a second point.
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What you'll master
Meet the destination, bring back what you already know, and gather the terms and formulas this lesson leans on.
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What you'll master
Know
Key facts
The $x$-intercept is found by setting $y = 0$; the $y$-intercept by setting $x = 0$.
Two points determine a line, so plotting both intercepts is usually enough to graph it.
Two lines are parallel exactly when their gradients are equal.
Two lines are perpendicular exactly when $m_1 m_2 = -1$, so each gradient is the negative reciprocal of the other.
Understand
Concepts
Why setting one variable to zero finds the intercept on the other axis.
Why a line through the origin needs a different graphing strategy.
Why the perpendicular condition is a negative reciprocal rather than just a negative.
Can do
Skills
Find both intercepts from any linear equation and use them to graph the line.
Choose an appropriate graphing technique, including for lines through the origin and for horizontal and vertical lines.
Find the equation of a line parallel or perpendicular to a given line through a given point.
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Key terms
$x$-interceptWhere the line crosses the $x$-axis, found by setting $y = 0$. Like this: in $y = 2x + 4$, setting $y = 0$ gives $x = -2$, so the $x$-intercept is $(-2, 0)$.
$y$-interceptWhere the line crosses the $y$-axis, found by setting $x = 0$. Like this: in $y = 2x + 4$, setting $x = 0$ gives $y = 4$.
Parallel linesLines with the same gradient, which never meet. Like this: $y = 3x + 1$ and $y = 3x - 7$ are parallel because both have gradient 3.
Perpendicular linesLines meeting at a right angle, whose gradients multiply to $-1$. Like this: $y = 2x$ and $y = -\frac{1}{2}x$ are perpendicular because $2 \times -\frac{1}{2} = -1$.
Negative reciprocalFlip a fraction and change its sign, which is how you get a perpendicular gradient. Like this: the negative reciprocal of $\frac{3}{4}$ is $-\frac{4}{3}$.
Horizontal and vertical linesLines of the form $y = k$ and $x = k$, which are perpendicular to each other but do not satisfy $m_1m_2 = -1$. Like this: $y = 3$ and $x = 5$ meet at a right angle, yet the vertical line has no gradient.
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Intercepts, and graphing from them
Work through the core explanation before applying it.
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Intercepts, and graphing from them
core concept
To find where a line crosses an axis, set the other variable to zero. For $2x + 3y = 12$: setting $y = 0$ gives $x = 6$, and setting $x = 0$ gives $y = 4$.
So the line passes through $(6, 0)$ and $(0, 4)$. Plot those two points and rule a line through them. Two points is all a straight line needs.
This method works directly from general form without rearranging, which is why it is usually the fastest route when an equation arrives as $ax + by = c$.
Label both intercepts on your sketch. Marks in graphing questions are usually for the intercepts, not for the neatness of the line.
Quick check: what is the $x$-intercept of $3x - 4y = 12$?
Set $y = 0$ for the $x$-intercept and $x = 0$ for the $y$-intercept, then plot both and join them. This works straight from general form without rearranging, and the intercepts are what earn the marks.
Pause, copy the two-intercept method with the worked $2x + 3y = 12$ giving $(6,0)$ and $(0,4)$, into your book.
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Choosing a graphing technique
core concept
We just saw that two intercepts usually give you the line. That raises a question: what about lines where that fails, such as one through the origin? This card answers it → pick the technique from the shape of the equation, which is exactly what the dot-point asks.
**Through the origin**, as in $y = 3x$, both intercepts are the same point $(0,0)$, so you have only one point. Plot the origin, then use the gradient to step to a second point: right 1, up 3.
**Horizontal or vertical**, as in $y = 4$ or $x = -2$. No calculation is needed; these are ruled straight across or straight up.
**Gradient-intercept form**, as in $y = \tfrac{1}{2}x - 3$. Plot the $y$-intercept, then step out by the gradient. Often faster than finding a fractional $x$-intercept.
Match the method to the equation. The dot-point says "choose and apply appropriate techniques", so a question may award a mark for picking the efficient route rather than grinding through intercepts every time.
Which line can NOT be graphed using two distinct intercepts?
Through the origin: plot $(0,0)$ and step out by the gradient. Horizontal or vertical: rule it directly. Gradient-intercept form: plot the intercept and step by the gradient. General form: use both intercepts. Choose from the shape of the equation.
Pause, copy the four cases and which technique suits each, and the worked step-out for $y = 3x$, into your book.
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Parallel and perpendicular
core concept
We just saw how to draw a single line. That raises a question: how do you tell from the equations alone whether two lines never meet, or meet at a right angle? This card answers it → compare gradients, since gradient is what fixes direction.
Two lines are **parallel** when their gradients are equal. $y = 3x + 1$ and $y = 3x - 7$ never meet: same direction, different starting height.
Two lines are **perpendicular** when $m_1 m_2 = -1$. Equivalently, each gradient is the negative reciprocal of the other: flip the fraction and change the sign. From $\tfrac{3}{4}$ you get $-\tfrac{4}{3}$.
To find a line through a given point parallel or perpendicular to a given one: get the given gradient, adjust it (keep it, or take the negative reciprocal), then use the point-gradient formula.
The one exception. $y = 3$ and $x = 5$ are perpendicular, but the vertical line has no gradient, so $m_1m_2 = -1$ cannot be tested. Horizontal and vertical lines are perpendicular by inspection.
Fill the blank: a line perpendicular to one with gradient $\tfrac{2}{5}$ has gradient $-\tfrac{5}{}$ where the missing denominator is .
Parallel means equal gradients. Perpendicular means $m_1m_2 = -1$, so each gradient is the negative reciprocal of the other: flip and change sign. To build such a line through a point, adjust the gradient then use the point-gradient formula. Horizontal and vertical lines are the exception, being perpendicular without a testable product.
Pause, copy both conditions, the negative-reciprocal procedure with $\frac{3}{4}$ giving $-\frac{4}{3}$, and the horizontal-vertical exception, into your book.
Worked examples · 3 in a row, reveal as you go
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Work examples end to end
Follow the reasoning through complete worked solutions.
PROBLEM 1 · GRAPHING FROM INTERCEPTS
Graph $2x + 3y = 12$ by finding both intercepts.
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Set $y = 0$: $2x = 12$, so $x = 6$
The $x$-intercept is $(6, 0)$.
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Set $x = 0$: $3y = 12$, so $y = 4$
The $y$-intercept is $(0, 4)$.
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Plot $(6,0)$ and $(0,4)$ and rule a line through them
Label both intercepts; they carry the marks.
PROBLEM 2 · PARALLEL THROUGH A POINT
Find the equation of the line parallel to $y = 4x - 7$ passing through $(2, 3)$.
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Parallel means the same gradient, so $m = 4$
Read the gradient off the given equation.
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$y - 3 = 4(x - 2)$
Point-gradient formula with the new point.
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$y = 4x - 5$
Expand and simplify. Check $(2,3)$: $8 - 5 = 3$ ✓
PROBLEM 3 · PERPENDICULAR THROUGH A POINT
Find the equation of the line perpendicular to $3x + 4y = 8$ passing through $(3, -1)$, in general form.
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Rearrange: $4y = -3x + 8$, so $y = -\tfrac{3}{4}x + 2$ and $m_1 = -\tfrac{3}{4}$
Get the given gradient first.
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Perpendicular gradient is the negative reciprocal: $m_2 = \tfrac{4}{3}$
Substitute, clear the fraction, then move to general form.
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Quick-fire practice
Work through the core explanation before applying it.
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Quick-fire practice
+10 XP
Find both intercepts of $5x - 2y = 20$.
Are $y = 2x + 1$ and $2y = 4x - 6$ parallel? Justify.
State the gradient of any line perpendicular to $y = -\tfrac{1}{3}x + 2$.
Find the line parallel to $y = -x + 4$ through $(0, 7)$.
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Revisit your perpendicular sketch
Run the quick drill and copy the summary into your book.
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Revisit your perpendicular sketch
At the start you sketched a line through the origin that looked perpendicular to $y = 2x + 4$, and guessed its gradient. Check your guess against $m_1m_2 = -1$ and write the equation of that line.
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Multiple choice
Answer the drill bank and rate your confidence.
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Multiple choice
+5 XP per correct · +25 XP all-correct
Pick your answer, then rate your confidence, that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
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Short answer
Write full responses, then check them against the model answers.
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Short answer
ApplyBand 43 marks
Q1. Find the $x$-intercept and $y$-intercept of $4x - 5y = 20$, and sketch the line, labelling both intercepts. (3 marks)
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ApplyBand 54 marks
Q2. Find the equation of the line perpendicular to $2x + 5y - 10 = 0$ that passes through $(4, 1)$. Give your answer in general form. (4 marks)
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UnderstandBand 42 marks
Q3. Explain why the perpendicular condition is that the gradients multiply to $-1$, rather than simply that one gradient is the negative of the other. (2 marks)
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📖 Comprehensive answers (click to reveal)
Practice 1: $x$-intercept $(4,0)$, $y$-intercept $(0,-10)$. Practice 2: yes, $2y = 4x - 6$ gives $y = 2x - 3$, gradient 2 in both, so they are parallel. Practice 3: 3, the negative reciprocal of $-\frac{1}{3}$. Practice 4: $y = -x + 7$.
Q1 (3 marks): Setting $y = 0$: $4x = 20$, so $x = 5$, giving $(5, 0)$ [1]. Setting $x = 0$: $-5y = 20$, so $y = -4$, giving $(0, -4)$ [1]. Line ruled through both points with each intercept labelled [1].
Q3 (2 marks): If $m_2$ were simply $-m_1$, the second line would be the mirror image of the first in the horizontal, which is generally not a right angle: $y = 2x$ and $y = -2x$ meet at an angle that is not $90^\circ$ [1]. Rotating a line through $90^\circ$ both reverses the sign and inverts the ratio of rise to run, because rise and run swap roles, which is exactly the negative reciprocal and gives $m_1m_2 = -1$ [1].
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Review and finish
Take the module quiz if you are ready, then mark the lesson complete.
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