Let \((x_1,y_1)=(-1,5)\) and \((x_2,y_2)=(3,-7)\). Then substitute into the gradient formula:

$$ \begin{eqnarray} m\ &=&\ \frac{y_2-y_1}{x_2-x_1}\\[6pt] &=&\ \frac{-7-5}{3--1}\\[6pt] &=&\ \frac{-12}{4}\\[6pt] &=&\ -\!3 \end{eqnarray} $$

Note that if we had taken these points the other way around, with \((x_1,y_1)=(3,-7)\) and \((x_2,y_2)=(-1,5)\), the formula would have given the same gradient.

Let \((x_1,y_1)=(2,-5)\) and \((x_2,y_2)=(-2,-11)\). Then substitute into the gradient formula:

$$ \begin{eqnarray} m\ &=&\ \frac{y_2-y_1}{x_2-x_1}\\[6pt] &=&\ \frac{-11--5}{-2-2}\\[6pt] &=&\ \frac{-11+5}{-4}\\[6pt] &=&\ \frac{-6}{-4}\\[6pt] &=&\ \frac{3}{2} \end{eqnarray} $$

Gradients that work out to be fractions should be left as fractions, not converted to decimals. Improper fractions should be left that way, not converted to mixed numbers. This is especially important when working within the straight line topic, where the working to find the equation of a straight line needs the gradient to be either a proper or improper fraction.

Let \((x_1,y_1)=(1,4)\) and \((x_2,y_2)=(-6,4)\). Then substitute into the gradient formula:

$$ \begin{eqnarray} m\ &=&\ \frac{y_2-y_1}{x_2-x_1}\\[6pt] &=&\ \frac{4-4}{-6-1}\\[6pt] &=&\ \frac{0}{-7}\\[6pt] &=&\ 0 \end{eqnarray} $$

Note that this is a horizontal line. We can see this without calculation from the fact that the two points have equal \(y\)-coordinates.

Let \((x_1,y_1)=(-3,-4)\) and \((x_2,y_2)=(-3,0)\). Then substitute into the gradient formula:

$$ \begin{eqnarray} m\ &=&\ \frac{y_2-y_1}{x_2-x_1}\\[6pt] &=&\ \frac{0--4}{-3--3}\\[6pt] &=&\ \frac{0+4}{-3+3}\\[6pt] &=&\ \frac{4}{0} \end{eqnarray} $$

Division by zero is impossible, so the gradient is undefined.

Note that this is a vertical line. We can see this without calculation from the fact that the two points have equal \(x\)-coordinates.

AB and CD are parallel if \(m_{\textsf{AB}}=m_{\textsf{CD}}\small.\) So we need to find each gradient and show that they are equal.

$$ \begin{eqnarray} m_{\textsf{AB}}\ &=&\ \frac{y_2-y_1}{x_2-x_1}\\[6pt] &=&\ \frac{-7-5}{1-(-2)}\\[6pt] &=&\ \frac{-12}{3}\\[6pt] &=&\ -\!4\\[15pt] m_{\textsf{CD}}\ &=&\ \frac{-10-(-2)}{7-5}\\[6pt] &=&\ \frac{-8}{2}\\[6pt] &=&\ -\!4\\[12pt] \end{eqnarray} $$

We have shown that \(m_{\textsf{AB}}=m_{\textsf{CD}}=-4\small,\) so AB and CD are indeed parallel.

Let \((x_1,y_1)=(1,5)\) and \((x_2,y_2)=(-2,p)\). Then substitute into the gradient formula:

$$ \begin{eqnarray} m\ &=&\ \frac{y_2-y_1}{x_2-x_1}\\[6pt] 4&=&\ \frac{p-5}{-2-1}\\[6pt] 4&=&\ \frac{p-5}{-3}\\[6pt] -12&=&\ p-5\\[6pt] p&=&\ -\!12+5\\[6pt] p&=&\ -\!7 \end{eqnarray} $$

Let \((x_1,y_1)=(1,a)\) and \((x_2,y_2)=(a,-1)\). Then substitute into the gradient formula:

$$ \begin{eqnarray} m\ &=&\ \frac{y_2-y_1}{x_2-x_1}\\[6pt] 2&=&\ \frac{-1-a}{a-1}\\[6pt] 2(a-1)&=&\ -1-a\\[6pt] 2a-2 &=&\ -1-a\\[6pt] 2a+a &=&\ -1+2\\[6pt] 3a&=&\ 1\\[6pt] a&=&\ \frac13 \\[6pt] \end{eqnarray} $$

This past paper question also involves factorising, including the difference of two squares, to simplify an algebraic fraction.

$$ \begin{eqnarray} m\ &=&\ \frac{y_2-y_1}{x_2-x_1}\\[6pt] &=&\ \frac{4p^2-9}{4p-6}\\[6pt] &=&\ \frac{(2p)^2-3^2}{2(2p-3)}\\[6pt] &=&\ \frac{(2p+3)(2p-3)}{2(2p-3)}\\[6pt] &=&\ \frac{2p+3}{2}\\[6pt] &=&\ p+\small\frac32 \end{eqnarray} $$

Either of the last two lines would be acceptable as your final answer.