Parallel & Perpendicular Lines (Cambridge (CIE) IGCSE Additional Maths): Revision Note

• Look for hidden parallel lines in an exam question

  • Parallel lines could be implied in an exam by phrases like “… at the same rate …”

  • Check the properties of a geometrical shape

Find the equation of the line that is parallel to $y=3x+7$ and passes through (2,1).

As the gradient is the same, the line that is parallel will be in the form: 

$y=3x+d$

Substitute in the coordinate that the line passes through: 

$1=3\left(2\right)+d$

Simplify: 

$1=6+d$

Subtract 6 from both sides: 

$-5=d$

Final answer: 

$\mathit{y}\mathbf{=}\mathbf{3}\mathit{x}\mathbf{-}\mathbf{5}$

• Exam questions are good at “hiding” perpendicular lines

  • For example a tangent and a radius are perpendicular

Find the equation of the line that is perpendicular to $y=2x-2$ and passes through (2, -3).

Leave your answer in the form $ax+by+c=0$ where $a,b,c$ are integers.  

L is in the form $y=mx+c$ so we can see that its gradient is 2

$m_1=2$

Therefore the gradient of the line perpendicular to L will be the negative reciprocal of 2

$m_2=-\frac{1}{2}$

Now we need to find $c$ for the line we're after Do this by substituting the point $\left(2,-3\right)$ into the equation $y=-\frac{1}{2}x+c$ and solving for $c$

$\begin{array}{rcl}-3& =& -\frac{1}{2}\times 2+c\\ -3& =& -1+c\\ c& =& -2\end{array}$

Now we know the line we want is 

$\begin{array}{rcl}y& =& -\frac{1}{2}x-2\end{array}$

But this is not in the form asked for in the question. So rearrange into the form $ax+by+c=0$ where $a$, $b$ and $c$ are integers

$\begin{array}{rcl}y+\frac{1}{2}x+2& =& 0\\ 2y+x+4& =& 0\end{array}$

Write the final answer

$\mathit{x}\mathbf{+}\mathbf{2}\mathit{y}\mathbf{+}\mathbf{4}\mathbf{=}\mathbf{0}$