Straight lines in coordinate geometry

A straight line has a constant gradient. Its equation records two pieces of information: how steeply the line rises or falls, and where it is positioned in the coordinate plane.

Straight-line methods support later work on circles, tangents, vectors, calculus and mathematical modelling. The formulae are short, but accurate reasoning about signs, vertical lines and context matters.

Before you begin

You should be able to:

• substitute signed numbers into an expression
• rearrange linear equations
• solve simultaneous linear equations
• use Pythagoras’ theorem

Gradient between two points

For distinct points $(x_1,y_1)$ and $(x_2,y_2)$ with $x_1\ne x_2$, the gradient is

$\boxed{m=\frac{y_2-y_1}{x_2-x_1}}.$

Gradient measures change in $y$ per unit change in $x$:

$m=\frac{\text{vertical change}}{\text{horizontal change}}.$

Use the same subtraction order in the numerator and denominator. Reversing both differences gives the same result; reversing only one changes the sign incorrectly.

Example 1: calculate a gradient

Find the gradient of the line through $A(-3,5)$ and $B(7,-1)$.

$m=\frac{-1-5}{7-(-3)}=\frac{-6}{10}=-\frac35.$

The negative sign agrees with the geometry: as $x$ increases from $-3$ to $7$, $y$ decreases.

If $x_1=x_2$, the denominator is zero and the line is vertical. Its gradient is undefined and its equation has the form

$\boxed{x=k}.$

A horizontal line has gradient $0$ and equation $y=k$.

Forms of a line equation

Different forms make different information visible.

Gradient-intercept form

$\boxed{y=mx+c}$

Here $m$ is the gradient and $c$ is the $y$ intercept, the value of $y$ when $x=0$.

Point-gradient form

A line of gradient $m$ through $(x_1,y_1)$ is

$\boxed{y-y_1=m(x-x_1)}.$

This is usually the quickest form when a point and gradient are known. It also avoids finding $c$ as a separate step.

General form

$\boxed{Ax+By+C=0}$

where $A$ and $B$ are not both zero. If $B\ne0$, rearranging gives

$y=-\frac ABx-\frac CB,$

so the gradient is $-A/B$. General form also includes vertical lines, for which $B=0$.

All three forms can describe the same line. For example,

$y-4=2(x-1),$

$y=2x+2,$

and

$2x-y+2=0$

are equivalent.

Forming an equation

Example 2: a point and a gradient

Find the equation of the line through $(4,-2)$ with gradient $3/2$.

Substitute into point-gradient form:

$y-(-2)=\frac32(x-4).$

Therefore

$\boxed{y+2=\frac32(x-4)}.$

If gradient-intercept form is required,

$y=\frac32x-8.$

Check the point: substituting $x=4$ gives $y=-2$.

Example 3: a line through two points

Find the equation through $P(-2,1)$ and $Q(4,10)$.

First find the gradient:

$m=\frac{10-1}{4-(-2)}=\frac96=\frac32.$

Use either point:

$y-1=\frac32(x+2).$

Hence

$\boxed{y=\frac32x+4}.$

Substituting both original points is a quick final check.

Parallel lines

Distinct parallel non-vertical lines have equal gradients:

$\boxed{m_1=m_2}.$

Vertical lines are also parallel to one another, although their gradients are undefined.

Example 4: a parallel line

Find the line through $(3,-5)$ parallel to

$4x-2y+7=0.$

Rearrange the given line:

$-2y=-4x-7,$

$y=2x+\frac72.$

Its gradient is $2$. The required line is

$\boxed{y+5=2(x-3)},$

or $y=2x-11$.

Alternatively, a parallel line in general form must have matching $x$ and $y$ coefficients up to a common factor, so it can be written $4x-2y+C=0$. Substituting $(3,-5)$ gives $C=-22$.

Perpendicular lines

For two non-vertical, non-horizontal perpendicular lines,

$\boxed{m_1m_2=-1}.$

Thus the perpendicular gradient to $m$ is

$\boxed{-\frac1m}.$

This is the negative reciprocal. Both operations matter.

Special cases must be handled geometrically:

• a horizontal line is perpendicular to a vertical line
• a vertical line is perpendicular to a horizontal line

Example 5: a perpendicular bisector

Find the perpendicular bisector of the segment joining $A(-1,2)$ and $B(5,6)$.

The midpoint is

$M=\left(\frac{-1+5}{2},\frac{2+6}{2}\right)=(2,4).$

The gradient of $AB$ is

$m_{AB}=\frac{6-2}{5-(-1)}=\frac23.$

The perpendicular gradient is $-3/2$. Therefore the perpendicular bisector is

$\boxed{y-4=-\frac32(x-2)}.$

This construction is important for finding the centre of a circle through known points.

Intersections

At an intersection, the coordinates satisfy both line equations. Solve the equations simultaneously.

Example 6: find an intersection

Find where

$2x+y=11$

and

$x-2y=-8$

intersect.

From the first equation,

$y=11-2x.$

Substitute into the second:

$x-2(11-2x)=-8,$

$5x-22=-8,$

$x=\frac{14}{5}.$

Then

$y=11-2\left(\frac{14}{5}\right)=\frac{27}{5}.$

The intersection is

$\boxed{\left(\frac{14}{5},\frac{27}{5}\right)}.$

Two distinct parallel lines have no intersection. Equivalent equations describe the same line and therefore have infinitely many common points.

Distance and midpoint

The distance between $(x_1,y_1)$ and $(x_2,y_2)$ follows from Pythagoras:

$\boxed{d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}.$

The midpoint is

$\boxed{M=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)}.$

Example 7: distance and midpoint

For $A(-4,1)$ and $B(2,9)$,

$$AB=\sqrt{(2-(-4))^2+(9-1)^2} =\sqrt{6^2+8^2}=10,$$

and

$M=\left(\frac{-4+2}{2},\frac{1+9}{2}\right)=(-1,5).$

Distance is never negative. Squaring coordinate differences ensures their signs do not affect the length.

Distance from a point to a line

For the line

$Ax+By+C=0,$

the perpendicular distance from $(x_0,y_0)$ is

$\boxed{d=\frac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}}}.$

The absolute value is essential because distance is non-negative. The denominator accounts for the scale of the equation: multiplying the whole line equation by a constant must not change the geometric distance.

Example 8: point-to-line distance

Find the distance from $P(3,-1)$ to

$4x-3y-6=0.$

Substitute into the formula:

$$d=\frac{|4(3)-3(-1)-6|}{\sqrt{4^2+(-3)^2}} =\frac{|9|}{5} =\boxed{\frac95}.$$

This formula is particularly useful for distances from circle centres to lines and for deciding whether a line meets a circle.

Straight-line models

A model of the form

$y=mx+c$

assumes a constant rate of change. In context:

• $m$ is the predicted change in $y$ for each one-unit increase in $x$
• $c$ is the predicted value of $y$ when $x=0$

The units belong in the interpretation. If $x$ is time in hours and $y$ is volume in litres, then $m$ is measured in litres per hour.

Example 9: interpret a model

The temperature $T$ in degrees Celsius of a cooling liquid is modelled for the first ten minutes by

$T=84-3.6t,$

where $t$ is time in minutes.

The intercept $84$ predicts an initial temperature of $84^\circ\text{C}$. The gradient $-3.6$ predicts that the temperature falls by $3.6^\circ\text{C}$ per minute.

At $t=7$,

$T=84-3.6(7)=58.8,$

so the model predicts $58.8^\circ\text{C}$.

Setting $T=0$ gives $t=84/3.6\approx23.3$ minutes, but this lies outside the stated ten-minute interval. It would also be physically unrealistic for a simple cooling process to continue linearly without limit.

Limitations of a linear model

Before using or extrapolating a straight-line model, ask:

• Is a constant rate plausible throughout the interval?
• Does the prediction remain within a physically meaningful range?
• Is the input inside the range from which the model was built?
• Does the intercept have a meaningful interpretation in context?
• Could another variable be affecting the relationship?

A good calculation can still produce a poor prediction if the model’s assumptions are unsuitable.

Common mistakes

• Mixing subtraction orders in the gradient formula.
• Reading $c$ as the $x$ intercept in $y=mx+c$.
• Assuming two lines are parallel because their intercepts are equal.
• Taking only the reciprocal, rather than the negative reciprocal, for a perpendicular gradient.
• Trying to assign a finite gradient to a vertical line.
• Finding the midpoint by subtracting coordinates instead of averaging them.
• Dropping the absolute value in the point-to-line distance formula.
• Giving a model gradient without its contextual units.
• Extrapolating a model far beyond its stated domain without comment.

Check your understanding

1. Find the equation of the line through $(-2,5)$ and $(4,-7)$.
2. Find the line through $(1,3)$ perpendicular to $3x+2y=8$.
3. Find the intersection of $y=3x-4$ and $2x+y=11$.
4. Find the perpendicular bisector of the segment joining $(2,-1)$ and $(8,3)$.
5. Find the distance from $(-2,4)$ to $5x+12y-9=0$.
6. A taxi fare is modelled by $C=3.20+1.85d$, where $C$ is cost in pounds and $d$ is distance in miles. Interpret both constants and predict the fare for $7$ miles.

Answers

1

$m=\frac{-7-5}{4-(-2)}=-2.$

Using $(-2,5)$,

$\boxed{y-5=-2(x+2)},$

or $y=-2x+1$.

2

Rearranging the given line gives $y=-3x/2+4$, so its gradient is $-3/2$. The perpendicular gradient is $2/3$:

$\boxed{y-3=\frac23(x-1)}.$

3

Substitute $y=3x-4$ into $2x+y=11$:

$2x+3x-4=11,$

so $x=3$ and $y=5$. The intersection is $\boxed{(3,5)}$.

4

The midpoint is $(5,1)$. The segment gradient is $4/6=2/3$, so the perpendicular gradient is $-3/2$:

$\boxed{y-1=-\frac32(x-5)}.$

5

$$d=\frac{|5(-2)+12(4)-9|}{\sqrt{5^2+12^2}} =\frac{29}{13}.$$

6

The fixed charge is £3.20 and the cost increases by £1.85 per mile. For $d=7$,

$C=3.20+1.85(7)=\boxed{£16.15}.$

What to learn next

Continue to circles to combine line equations with radii, chords and tangents. Then study parametric equations to represent curves and motion using a third variable.