$\begin{align}(i)&(4,6) (2,2)\\[5pt] &m=\frac{6-2}{4-2}=\frac{4}{2}\\[5pt] &=\underline{\underline{2}} \end{align}$	$\begin{align}(ii)&(6,2) (4,3)\\[5pt] &m=\frac{2-3}{6-4}\\[5pt] &=\underline{\underline{\frac{-1}{2}}} \end{align}$	$\begin{align}(iii)&(1,-2) (0,7)\\[5pt] &m=\frac{-2-7}{1-0}=\frac{-9}{1}\\[5pt] &=\underline{\underline{-9}} \end{align}$
$\begin{align}(iv)&(-2,-3) (2,5)\\[5pt] &m=\frac{5-(-3)}{2-(-2)}=\frac{5+3}{2+2}=\frac{8}{4}\\[5pt] &=\underline{\underline{2}} \end{align}$	$\begin{align}(v)&(4,5) (-8,-4)\\[5pt] &m=\frac{5-(-4)}{4-(-8)}=\frac{9}{12}\\[5pt] &=\underline{\underline{\frac{4}{3}}} \end{align}$	$\begin{align}(vi)&(6,-4) (2,2)\\[5pt] &m=\frac{-4-2}{6-2}=\frac{-6}{4}\\[5pt] &=\underline{\underline{\frac{-3}{2}}} \end{align}$
$\begin{align}(vii)&(1,-4) (-2,-7)\\[5pt] &m=\frac{-4-(-7)}{1-(-2)}=\frac{-4+7}{1+2}=\frac{3}{3}\\[5pt] &=\underline{\underline{1}} \end{align}$	$\begin{align}(viii)&(4,6) (-2,-9)\\[5pt] &m=\frac{6-(-9)}{4-(-2)}=\frac{15}{6}\\[5pt] &=\underline{\underline{\frac{5}{2}}} \end{align}$

By studying this lesson you will be able to

Find the gradient of the graph of a straight line.