3.5, 3.6 - Equations of straight lines

Connect a line's equation to its gradient, intercepts and points. In 3.5 you will interpret and construct equations in different forms, including parallel, perpendicular and vertical lines; in 3.6 you will use points and a gradient to draw the required portion of a line accurately.

Reading a line's gradient and intercept

The lines $y=\frac23x+\frac13$ and $y=-\frac23x+\frac53$ have different gradients and different intercepts. Can you spot a point they share without drawing them? The final part will reveal a whole family of lines through that same point.

An equation describes every point on a line: a coordinate pair is on the line precisely when substituting it makes the equation true. Rearranging the equation changes its appearance, but not the points that satisfy it.

Intercept

An intercept is where a graph meets an axis. At a $y$ -intercept, $x=0$ ; at an $x$ -intercept, $y=0$ . For example, $y=2x-3$ has $y$ -intercept $(0,-3)$ . The number $-3$ is its intercept height, not its gradient.

The form $y=mx+c$ separates a line's direction from its position. Increasing $x$ by $1$ increases $y$ by $m$ , while putting $x=0$ gives $y=c$ .

Gradient-intercept form

A non-vertical straight line can be written as

y=mx+c

Its gradient is $m$ and its $y$ -intercept is $(0,c)$ . A horizontal line has $m=0$ .
Not on the formulae sheet: learn it.

Other forms, such as $ax+by=d$ , often hide the gradient. Make $y$ the subject before reading it.

Worked example 1: a hidden positive gradient

Question. Work out the gradient and $y$ -intercept of $3x-2y=8$ .

Plan. Rearrange into $y=mx+c$ , dividing every term by the coefficient of $y$ .

\begin{aligned} -2y&=8-3x && \text{subtract }3x\\ y&=\frac32x-4 && \text{divide both terms by }-2 \end{aligned}

Why this step? The coefficient of $x$ gives the gradient only after $y$ stands alone with coefficient $1$ .

Answer. Gradient $\frac32$ ; $y$ -intercept $(0,-4)$ .

Check. Substituting $(0,-4)$ gives $0-2(-4)=8$ . Increasing $x$ by $2$ must increase $y$ by $3$ , which keeps $3x-2y$ unchanged.

Worked example 2 changes only the sign of the $2y$ term. Both the gradient and intercept change.

Worked example 2: a hidden negative gradient

Question. Work out the gradient and $y$ -intercept of $3x+2y=8$ .

Plan. Make $y$ the subject, retaining the sign of the moved $3x$ term.

\begin{aligned} 2y&=8-3x && \text{subtract }3x\\ y&=-\frac32x+4 && \text{divide every term by }2 \end{aligned}

Why this step? Moving $3x$ to the other side makes its coefficient negative, so this line falls from left to right.

Answer. Gradient $-\frac32$ ; $y$ -intercept $(0,4)$ .

Check. The point $(2,1)$ satisfies $3(2)+2(1)=8$ and $1=-\frac32(2)+4$ , confirming that both forms describe the same line.

A vertical line is the exception to $y=mx+c$ . Every point on $x=5$ has horizontal coordinate $5$ , but its height is unrestricted, so there is no defined gradient and no $y$ -intercept. A horizontal line such as $y=-2$ has gradient $0$ . The vertical line $x=0$ is the whole $y$ -axis, rather than meeting it at a single intercept.

More generally, $ax+by+d=0$ is a straight-line equation when $a$ and $b$ are not both zero. If $b\ne0$ , rearrange to $y=-\frac abx-\frac db$ ; if $b=0$ , it becomes the vertical line $x=-\frac da$ .

Watch out

Wrong: the gradient of $3x+2y=8$ is $3$ . Right: divide the rearranged equation by $2$ , giving gradient $-\frac32$ . Read the gradient after rearranging, and give the coefficient itself rather than a term containing $x$ .

Read the next equation in the same way, keeping the entire rearrangement exact.

Writing a line through a known point

A point fixes a line's position once its gradient fixes its direction.

You already know

For two points with different $x$ -coordinates, use $m=\frac{y_2-y_1}{x_2-x_1}$ . A point on the line must satisfy its equation. If two distinct points share the same $x$ -coordinate, their line is vertical instead.

Measure changes from a known point $(x_1,y_1)$ to a variable point $(x,y)$ on the line. Their gradient relation is $\frac{y-y_1}{x-x_1}=m$ , so multiplying through gives the point-gradient form. The known point also satisfies the multiplied form directly.

Point-gradient form

A line of finite gradient $m$ through $(x_1,y_1)$ has equation

y-y_1=m(x-x_1)

This form does not describe a vertical line; use $x=x_1$ for that case.
On the formulae sheet.

You may leave a correct equation in this form unless the question requests another form. Expand and collect only when useful.

Worked example 3: a point and a gradient

Question. Work out the equation of the line through $(2,-3)$ with gradient $2$ . Give your answer in the form $y=mx+c$ .

Plan. Substitute the known point into point-gradient form, then make $y$ the subject.

\begin{aligned} y-(-3)&=2(x-2) && \text{use the point and gradient}\\ y+3&=2x-4 && \text{expand the bracket}\\ y&=2x-7 && \text{subtract }3 \end{aligned}

Why this step? The differences are $y-(-3)$ and $x-2$ , so the given point makes both differences zero.

Answer. $y=2x-7$ .

Check. At $x=2$ , $y=4-7=-3$ , and the coefficient of $x$ is the required gradient $2$ .

Worked example 4 replaces the supplied gradient with a second point on the same line. The equation should therefore stay unchanged.

Worked example 4: two points supply the gradient

Question. Work out the equation of the line through $P(2,-3)$ and $Q(6,5)$ .

Plan. Calculate the gradient first, then use either point in point-gradient form.

\begin{aligned} m&=\frac{5-(-3)}{6-2}=2 && \text{the two points give the direction}\\ y+3&=2(x-2) && \text{use point }P\\ y&=2x-7 && \text{expand and rearrange} \end{aligned}

Why this step? Two distinct points determine one line. The gradient is the missing information that lets you use the same formula as before.

Answer. $y=2x-7$ , or any equivalent straight-line equation.

Check. At $Q$ , $2(6)-7=5$ ; at $P$ , $2(2)-7=-3$ . Both supplied points lie on the result.

You can instead substitute into $y=mx+c$ to work out $c$ . For the first example, $-3=2(2)+c$ gives $c=-7$ . This is the same method written in another form, not a different line.

Watch out

Wrong: $y-3=2(x-2)$ for the point $(2,-3)$ . It fails when that point is substituted. Right: $y+3=2(x-2)$ . Keep subtraction of a negative coordinate in brackets until the signs are settled.

Calculate a new gradient and give the resulting equation exactly.

Equations with a parallel or perpendicular condition

A relationship to another line supplies the gradient, while the given point supplies the position of your new line.

For a non-vertical reference line, first make $y$ the subject. Keep its gradient for a parallel direction, or take its negative reciprocal for a perpendicular direction. Then use the point you were actually given for the new line, not a convenient point on the reference line.

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Number

Algebra

Coordinate geometry

Calculus

Matrix transformations

Geometry

3.5, 3.6 - Equations of straight lines

Reading a line's gradient and intercept

Writing a line through a known point

Equations with a parallel or perpendicular condition

Continue this lesson.

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