National 5 Maths
Gradient

Let \((x_1,\,y_1)=(-1,\,5)\) and \((x_2,\,y_2)=(3,\,-7)\small.\)

Then substitute into the gradient formula:

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

Note 1: It doesn't matter which point is \((x_1,\,y_1)\) and which is \((x_2,\,y_2)\small.\) If we take these points the other way around, with \((x_1,\,y_1)=(3,\,-7)\) and \((x_2,\,y_2)=(-1,\,5)\small,\) the formula will give the same gradient.

Note 2: For the same reason, using \(m=\large\frac{y_1\,-\,y_2}{x_1\,-\,x_2}\) will also work perfectly.

Note 3: Just be careful not to 'mix and match'. \(m=\large\frac{y_2\,-\,y_1}{x_1\,-\,x_2}\) and \(m=\large\frac{y_1\,-\,y_2}{x_2\,-\,x_1}\) are both wrong.

Video solution by Clelland Maths 

Let \((x_1,\,y_1)=(2,\,-5)\) and \((x_2,\,y_2)=(-2,\,-11)\small.\)

Then substitute into the gradient formula:

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

Note: Gradients that work out to be fractions should be left as fractions, not converted to decimals. 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.

Video solution by Clelland Maths 

Let \((x_1,\,y_1)=(1,\,4)\) and \((x_2,\,y_2)=(-6,\,4)\small.\)

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: This is a horizontal line. We can see this from the fact that the points have the same \(y\)-coordinates.

Video solution by Clelland Maths 

Let \((x_1,\,y_1)=(-3,\,-4)\) and \((x_2,\,y_2)=(-3,z,0)\small.\)

Then substitute into the gradient formula:

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

Note 1: The reason this gradient is undefined is that \(4\div 0\) does not exist. Division by zero is impossible.

Note 2: This is a vertical line. We can see this from the fact that the two points have the same \(x\)-coordinates.

Video solution by Clelland Maths 

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.

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

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

Video solution by Clelland Maths 

We have been told that the gradient is \(4\small.\) So we have to work backwards to find the unknown \(y\)-coordinate.

Let \((x_1,\,y_1)=(1,\,5)\) and \((x_2,\,y_2)=(-2,\,p)\small.\)

Then substitute into the gradient formula:

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

Video solution by Clelland Maths 

This question is similar to example 6, but the unknown is in the coordinates of both points.

Let \((x_1,\,y_1)=(1,\,a)\) and \((x_2,\,y_2)=(a,\,-1)\small.\)

Then substitute into the gradient formula:

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

Video solution by Clelland Maths 

This 3-mark question needs us to factorise to simplify an algebraic fraction.

Note that the numerator is a difference of two squares and the denominator has a common factor.

Once the numerator and denominator are both fully factorised, we cancel the \((2p-3)\) top and bottom.

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

Note: Either of the last two lines is acceptable as the final answer.

Video solution by Clelland Maths