Straight linesAQA A-Level Maths: Revision notes

The gradient of the line through $(x_1,y_1)$ and $(x_2,y_2)$ is $m=\frac{y_2-y_1}{x_2-x_1}$. In the form $y=mx+c$, $m$ is the gradient and $c$ is the $y$-intercept. For $A(1,2)$ and $B(5,12)$, $m=\frac{10}{4}=\frac52$. A positive gradient slopes upwards from left to right, a negative gradient slopes downwards, and the gradient of a horizontal line is $0$. Prior GCSE knowledge is assumed for the midpoint $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$ and the distance between two points.

Given the gradient $m$ and one point $(x_1,y_1)$, the equation is $y-y_1=m(x-x_1)$. Given two points, find $m$ first and then use either point. Example: through $A(1,2)$ and $B(5,12)$: $y-2=\frac52(x-1)$, so $2y-4=5x-5$ and $y=\frac52x-\frac12$. A point lies on a line if its coordinates satisfy the equation. The line meets the $y$-axis where $x=0$ and the $x$-axis where $y=0$.

Many questions ask for the answer as $ax+by+c=0$ with $a$, $b$, $c$ integers. Multiply through to clear fractions and move every term to one side. From $y=\frac52x-\frac12$: multiply by 2 to get $2y=5x-1$, so $5x-2y-1=0$. To find the gradient of a line given in this form, rearrange: $by=-ax-c$, so $m=-\frac{a}{b}$. For $3x+4y-12=0$: $m=-\frac34$.

Parallel lines have equal gradients: $m_1=m_2$. Perpendicular lines have gradients whose product is $-1$: $m_1m_2=-1$, so $m_2=-\frac{1}{m_1}$ (flip the fraction and change the sign). A line parallel to $3x+4y-12=0$ through $(8,-1)$ has $m=-\frac34$: $y+1=-\frac34(x-8)$, so $3x+4y-20=0$. The perpendicular bisector of $AB$ passes through the midpoint of $AB$ with gradient $-\frac{1}{m_{AB}}$. For $A(2,1)$ and $B(6,3)$: midpoint $(4,2)$, $m_{AB}=\frac12$, so the bisector is $y-2=-2(x-4)$, i.e. $2x+y-10=0$.

A linear model $y=mx+c$ describes a situation with a constant rate of change. The gradient is the change in the output per unit increase in the input, and the intercept is the starting value (when the input is zero). Example: a taxi costing £9.50 for 4 km and £20 for 10 km has $m=\frac{10.5}{6}=1.75$ (£ per km) and $C=2.5+1.75d$, so £2.50 is the fixed charge. Two models can be compared by solving their equations simultaneously, and the one with the smaller gradient is cheaper beyond the crossing point. A linear model may fail for values far outside the data, because real rates may change.