8D - Equations of straight lines

A point fixes a position; a gradient fixes a direction. You will combine them to describe a whole line, recognise equivalent equation forms and find where lines meet by solving their equations together.

A point and a gradient

How can one point and one gradient determine every other point on a line?

Take a line through $(2,5)$ with gradient $3$ . Moving horizontally from $2$ to $x$ changes $x$ by $x-2$ , so the matching vertical change must be $3(x-2)$ . The new vertical coordinate therefore satisfies $y-5=3(x-2)$ .

Point-gradient form

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

y-y_1=m(x-x_1).

Every point on the line satisfies this relation, including the given point, where both changes are zero.

Not on the formulae sheet: learn it.

The equation compares coordinate changes. This explains why both brackets must refer to the same starting point.

Worked example 1: using a positive gradient

Question. Find the equation of the line through $(2,5)$ with gradient $3$ , in the form $y=mx+c$ .

Plan. Substitute the point and gradient, then expand and isolate $y$ .

\begin{aligned} y-5&=3(x-2) &&\text{use the given point}\\ y-5&=3x-6 &&\text{expand the bracket}\\ y&=3x-1 &&\text{add }5 \end{aligned}

Answer. $y=3x-1$ .

Check. At $x=2$ , the equation gives $y=5$ . Increasing $x$ by $1$ increases $y$ by $3$ , so the gradient is correct too.

Now use a negative fractional gradient and a negative starting coordinate. The method is unchanged, but both signs need care.

Worked example 2: a negative fractional gradient

Question. Find the equation through $(-2,5)$ with gradient $-\tfrac32$ , in the form $y=mx+c$ .

Plan. Write $x-(-2)=x+2$ before expanding.

\begin{aligned} y-5&=-\frac32(x+2)\\ y-5&=-\frac32x-3\\ y&=-\frac32x+2 \end{aligned}

Answer. $y=-\tfrac32x+2$ .

Check. Substituting $x=-2$ gives $3+2=5$ . From $(-2,5)$ to the point $(0,2)$ , the gradient is $(2-5)/(0-(-2))=-\tfrac32$ .

One coordinate alone cannot check a proposed equation: the given point and the gradient must both agree.

Watch out

Through $(-2,5)$ , the bracket is $x+2$ , not $x-2$ . Substituting the given point into the completed equation is a quick way to catch this sign error.

Keep the fraction exact in the next equation.

Equivalent line forms

Rearranging a line's equation changes its appearance while keeping the same set of points.

For a nonvertical line, $y=mx+c$ displays its gradient $m$ and its $y$ -intercept $(0,c)$ . An intercept is a point where a line meets a coordinate axis. Set $y=0$ for an $x$ -intercept, and $x=0$ for a $y$ -intercept.

[DIAGRAM: asset_name: Intercepts and a vertical line; asset_slug: line-forms; caption: The line 3x plus 2y equals 12 has two axis intercepts, while x equals 4 is vertical; description: Two equal-scale panels show the line 3x plus 2y equals 12 through 0 comma 6 and 4 comma 0, and the vertical line x equals 4. The vertical line crosses the x-axis at 4 comma 0 and never crosses the y-axis.]

General straight-line form

ax+by=c

describes a straight line when $a$ and $b$ are not both zero.

If $b\ne0$ , rearranging gives

y=-\frac abx+\frac cb.

If $b=0$ , then $a\ne0$ and the line is vertical: $x=c/a$ .

Not on the formulae sheet: learn these forms.

A horizontal line has equation $y=k$ . A vertical line has equation $x=k$ and cannot be written with a finite gradient in the form $y=mx+c$ . The exceptional case $a=b=0$ would give either all points or no points, rather than a line.

Worked example 3: reading gradient and intercepts

Question. Find the gradient and both axis intercepts of $3x+2y=12$ .

Plan. Isolate $y$ for the gradient, then set one coordinate to zero for each intercept.

\begin{aligned} 2y&=12-3x &&\text{subtract }3x\\ y&=-\frac32x+6 &&\text{divide every term by }2 \end{aligned}

With $y=0$ , $3x=12$ , so $x=4$ . With $x=0$ , $2y=12$ , so $y=6$ .

Answer. Gradient $-\tfrac32$ ; intercepts $(4,0)$ and $(0,6)$ .

Check. Both intercepts satisfy the original equation, and their gradient is $(0-6)/(4-0)=-\tfrac32$ .

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8D - Equations of straight lines

A point and a gradient

Equivalent line forms

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