A9 - Equations of straight lines | AQA GCSE Mathematics

A gradient tells you a line's direction, but one more piece of information fixes its position. You will use parallel lines, a point and gradient, or two points to work out a straight-line equation. The final part uses perpendicular gradients at Higher tier.

Parallel lines and their gradients

Two lines both rise $3$ units for every unit right. One passes through $(0,1)$ , the other through $(0,-4)$ . Will they ever meet, however far you extend them? Their equations will settle the question.

You already know

For a non-vertical line in the form $y=mx+c$ , $m$ is its gradient and $(0,c)$ its $y$ -intercept. Rearrange so that $y$ has coefficient $1$ before reading these values.

Compare $y=3x+1$ and $y=3x-4$ . At any shared $x$ -value, subtracting their heights gives $(3x+1)-(3x-4)=5$ . The gap never closes. That constant gap is why the lines cannot meet; it is not just how the picture looks.

[DIAGRAM: asset_name: Parallel lines keep a fixed separation; asset_slug: aqa8300_a19_parallel; caption: Equal gradients keep the two lines apart at every x-value; description: Equal-scale coordinate axes showing y equals 3x plus 1 and y equals 3x minus 4. Their different y-intercepts are labelled and vertical segments at x equals 0 and x equals 1 each show a height difference of 5.]

Parallel gradients

Distinct non-vertical straight lines are parallel exactly when their gradients are equal:

m_1=m_2

In $y=m_1x+c_1$ and $y=m_2x+c_2$ , equal gradients and equal intercepts give the same line. All vertical lines are parallel to one another.

Not on the formulae sheet: learn it.

A different-looking equation can still describe the same line. Simplify before deciding whether there are two distinct lines.

Worked example 1: matching gradients

Question. Which of these lines is parallel to $y=3x+1$ ? Non-calculator.

A:\ y=3x-4\qquad B:\ y=-3x+1\qquad C:\ y=x+3

Plan. Compare the coefficient of $x$ in each equation, with $y$ as the subject.

\begin{aligned} m_{\text{given}}&=3 && \text{read the gradient}\\ m_A&=3,\quad m_B=-3,\quad m_C=1 && \text{compare gradients} \end{aligned}

Why this step? Parallel lines have the same rate of rise; the intercept only moves the line's position.

Answer. Line $A$ .

Check. $A$ 's intercept is $-4$ , different from $1$ , so it is a distinct line with the same gradient.

Change only the form of the given equation. Its gradient is hidden until you rearrange.

Worked example 2: writing a new parallel line

Question. Write down the equation of a line parallel to $2y=6x+2$ . Non-calculator.

Plan. Divide by $2$ and keep the gradient, choosing a different intercept.

\begin{aligned} y&=3x+1 && \text{divide every term by }2\\ m&=3 && \text{read the gradient}\\ y&=3x+4 && \text{choose a different intercept} \end{aligned}

Why this step? A new intercept gives a different line; simply rewriting the given equation does not.

Answer. For example, $y=3x+4$ .

Check. The height difference is always $3$ , so the two lines cannot meet.

There are infinitely many possible answers to “a line parallel to”, unless the question also fixes a point or intercept. Write one actual equation with a numerical constant.

Watch out

Wrong: answering $y=3x+1$ for a parallel line to $2y=6x+2$ . This is the same line. Right: keep $3$ but change the intercept, for example $y=3x-2$ .

Try a new coefficient and make sure your answer names one distinct line.

A line through one point with a given gradient

A gradient fixes the slope. A point fixes which of the parallel lines you want. Start with $y=mx+c$ , put in the known gradient, then substitute the point's two coordinates to work out $c$ .

For gradient $2$ , all candidates look like $y=2x+c$ . Requiring the line to pass through $(3,8)$ gives $8=2(3)+c$ , leaving just $c$ unknown. Subtracting $6$ from both sides fixes the intercept.

Finding the intercept from a point

For a line of gradient $m$ through $(x_0,y_0)$ ,

c=y_0-mx_0

This is $y=mx+c$ rearranged. It applies to non-vertical lines.

Not on the formulae sheet: learn it.

You can use the original equation directly instead of memorising another arrangement. These examples show the substitution in full.

Worked example 3: a point with positive coordinates

Question. Work out the equation of the line with gradient $2$ through $(3,8)$ . Non-calculator.

Plan. Substitute the point into $y=2x+c$ , solve for $c$ , then write the whole equation.

\begin{aligned} y&=2x+c && \text{use the gradient}\\ 8&=2(3)+c && \text{substitute both coordinates}\\ c&=8-6=2 && \text{subtract }6 \end{aligned}

Why this step? A point on the line must make its equation true, so it supplies an equation for the missing constant.

Answer. $y=2x+2$ .

Check. At $x=3$ , $2(3)+2=8$ , and the coefficient of $x$ remains $2$ .

Change only the sign of the point's $x$ -coordinate. Keep that negative inside brackets.

Worked example 4: a negative coordinate

Question. Work out the equation of the line with gradient $2$ through $(-3,8)$ . Non-calculator.

Plan. Substitute into the same family $y=2x+c$ .

\begin{aligned} 8&=2(-3)+c && \text{substitute the signed coordinate}\\ 8&=-6+c && \text{multiply}\\ c&=14 && \text{add }6\text{ to both sides} \end{aligned}

Why this step? The point is three units left of the $y$ -axis. Moving right to the intercept raises the line by $6$ , so its intercept must be higher than $8$ .

Answer. $y=2x+14$ .

Check. $2(-3)+14=8$ .

This method also finds a parallel line through a given point: first copy the other line's gradient, then substitute your point.

Watch out

Wrong: using $c=8$ just because the point is $(-3,8)$ . Right: $c$ is the height at $x=0$ , not the height at every point. Here substitution gives $c=14$ .

Your answer should finish with an equation, rather than just the value of $c$ .

A line through two points

Two points supply both ingredients: their changes give the gradient, then either point gives the intercept. Use the same order in the numerator and denominator of the gradient fraction.

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A9 - Equations of straight lines

Parallel lines and their gradients

A line through one point with a given gradient

A line through two points

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