Straight-Line Graphs

A coordinate pair $(x,y)$ locates a point on a grid. The horizontal axis is the $x$-axis; the vertical is the $y$-axis. They intersect at the origin $(0,0)$.

The axes divide the plane into four quadrants:

Quadrant	$x$	$y$	Example
1st (top right)	positive	positive	$(3,4)$
2nd (top left)	negative	positive	$(-2,5)$
3rd (bottom left)	negative	negative	$(-3,-1)$
4th (bottom right)	positive	negative	$(4,-2)$

Reading coordinates: the $x$-value (horizontal) is always stated first. A common error is reversing the pair.

To plot $(-3,4)$: move 3 left from the origin, then 4 up.

(Extra context — the midpoint formula belongs to G11 in the Edexcel 1MA1 spec, not A8. It appears here as a natural companion to coordinate reading and is assessed alongside straight-line graph work.)

Midpoint of two points $(x_1,y_1)$ and $(x_2,y_2)$: $\text{Midpoint}=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$

Worked example — midpoint of $(-2,6)$ and $(4,-2)$: $\left(\frac{-2+4}{2},\frac{6+(-2)}{2}\right)=(1,2)$ ✓

The gradient $m$ measures the steepness of a line — rise over run.

$m=\frac{\text{rise}}{\text{run}}=\frac{\text{change in }y}{\text{change in }x}=\frac{y_2-y_1}{x_2-x_1}$

• Positive gradient: line slopes upward left to right.
• Negative gradient: line slopes downward left to right.
• Zero gradient: horizontal line.
• Undefined gradient: vertical line.

Worked example — find the gradient of the line through $(1,3)$ and $(5,11)$:

$m=\frac{11-3}{5-1}=\frac{8}{4}=2$

Worked example — find the gradient of the line through $(-2,5)$ and $(4,2)$:

$m=\frac{2-5}{4-(-2)}=\frac{-3}{6}=-\frac{1}{2}$

Every straight line (except vertical lines) can be written as $y=mx+c$, where:

• $m$ = gradient (steepness and direction)
• $c$ = $y$-intercept (where the line crosses the $y$-axis; the value of $y$ when $x=0$)

Worked example — for the line $y=3x-2$:

• Gradient: $m=3$ (goes up 3 for every 1 right)
• $y$-intercept: $c=-2$ (crosses $y$-axis at $(0,-2)$)

Reading the equation from a graph: identify where the line crosses the $y$-axis (gives $c$) and pick two clear points to calculate the gradient (gives $m$).

Converting to $y=mx+c$:

$3x-2y=6⟹-2y=-3x+6⟹y=\frac{3}{2}x-3$

Gradient $=\frac{3}{2}$; $y$-intercept $=-3$. ✓

Given gradient $m$ and a point $(x_1,y_1)$ — use $y-y_1=m(x-x_1)$, then rearrange to $y=mx+c$.

Worked example — find the equation of the line with gradient 2 through $(3,7)$:

$y-7=2(x-3)⟹y-7=2x-6⟹y=2x+1$

Given two points — find the gradient first, then use $y-y_1=m(x-x_1)$.

Worked example — find the equation of the line through $(1,5)$ and $(4,14)$:

$m=\frac{14-5}{4-1}=\frac{9}{3}=3$

$y-5=3(x-1)⟹y=3x+2$

Check: when $x=4$: $y=12+2=14$ ✓

Parallel lines have the same gradient. If a line has gradient $m$, any parallel line also has gradient $m$.

Perpendicular lines cross at right angles. If a line has gradient $m$, a perpendicular line has gradient $-\frac{1}{m}$ (the negative reciprocal). Their gradients multiply to $-1$.

$m_1\times m_2=-1\text{(perpendicular)}$

Worked example — find the equation of the line perpendicular to $y=3x+1$ that passes through $(6,2)$.

Gradient of $y=3x+1$: $m=3$

Perpendicular gradient: $m_{\perp }=-\frac{1}{3}$

$y-2=-\frac{1}{3}(x-6)⟹y=-\frac{1}{3}x+2+2=-\frac{1}{3}x+4$

Check: at $(6,2)$: $-\frac{1}{3}(6)+4=-2+4=2$ ✓

Linear graphs in real contexts have practical meanings for gradient and intercept.

Distance-time graph: gradient = speed (distance ÷ time); a horizontal section means stationary; a steeper line means faster speed.

Conversion graphs (e.g. £ to $):gradientistheconversionrate;$y$-intercept is 0 if proportional.

Cost function — a plumber charges £50 call-out plus £30/hour. Cost $=30h+50$:

• Gradient (£30): additional cost per hour
• $y$-intercept (£50): fixed call-out charge, paid even if $h=0$

Worked example — a straight-line graph passes through $(0,80)$ and $(4,40)$.

Gradient: $m=\frac{40-80}{4-0}=-10$. Equation: $y=-10x+80$.

In context (e.g. water draining): the initial volume is 80 litres; it decreases at 10 litres per minute.

1. Reading $y=mx+c$ — confusing gradient and intercept

In $y=2x+5$, the gradient is 2 and the $y$-intercept is 5. In $y=-3x+2$, the gradient is $-3$ (negative) and the $y$-intercept is 2.

2. Calculating gradient — not using full coordinate differences

Gradient requires $\frac{y_2-y_1}{x_2-x_1}$. Reading off only a vertical distance without dividing by the horizontal distance gives the rise, not the gradient.

3. Perpendicular gradient — forgetting the negative

The perpendicular gradient is the negative reciprocal. The perpendicular to a line with gradient 4 has gradient $-\frac{1}{4}$, not $\frac{1}{4}$.

4. Line through two points — not checking the answer

After finding the equation, substitute both original points to verify. If either fails to satisfy the equation, the gradient calculation was wrong.

Mistake	Correction
"Parallel to $y=3x+1$, so gradient = $\frac{1}{3}$"	Parallel lines share the same gradient: gradient = 3
"Gradient of line through $(2,3)$ and $(5,9)$ is $\frac{6}{3}=2$"	$m=\frac{9-3}{5-2}=\frac{6}{3}=2$ — this one is correct; a common version of the error would be using $x_2-x_1$ in the numerator
"Perpendicular to $y=-2x+3$ has gradient 2"	Perpendicular gradient is $-\frac{1}{-2}=\frac{1}{2}$ (negative reciprocal, not just reciprocal)