Equations of a Straight Line (Edexcel IGCSE Further Maths)

Spec point — 8D

Equations of a Straight Line

How do I find the gradient of a straight line?

• Find two points that the line passes through with coordinates (x₁, y₁) and (x₂, y₂)
• The gradient m between these two points is calculated by

$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

• This is sometimes known as rise over run

• The gradient of a straight line measures its slope

  • A line with gradient 1 will go up 1 unit for every unit it goes to the right
  • A line with gradient -2 will go down two units for every unit it goes to the right

What are the equations of a straight line?

• $y=mx+c$

  • This is sometimes called gradient-intercept form
  • It clearly shows the gradient m and the y-intercept (0, c)

• $y-y_{1}=m(x-x_{1})$

  • This is sometimes called the point-gradient form
  • It clearly shows the gradient m and a point on the line (x₁, y₁)

• $ax+by=c$

  • This is sometimes called the general form

  • You can quickly get the x-intercept$(\frac{c}{a},0)$, y-intercept$(0,\frac{c}{b})$ and gradient $m=-\frac{a}{b}$

    • $c$ is not the *y-*intercept in this form of the line equation!

  • You can also rearrange the equation into gradient-intercept form

    • $y=-\frac{a}{b}x+\frac{c}{b}$

How do I find an equation of a straight line?

• You will need the gradient

  • If you are given two points then first find the gradient

• It is easiest to start with the point-gradient form$y - y_{1} = m ( x - x_{1} )$

  • then rearrange into whatever form is required

    • multiplying both sides by any denominators will get rid of fractions

• Always check your answer

  • Substitute the coordinates of points on the line into the equation
  • Make sure the equation is satisfied with those coordinates

Exam Hint

• Make sure you state equations of straight lines in the form required

  • Usually $y=mx+c$ or $ax+by=c$

  • Check whether coefficients need to be integers

    • This is often the case for $ax+by=c$

Worked Example The line $l$ passes through the points $(-2,5)$ and $(6,-7)$.

Find the equation of $l$ , giving your answer in the form $ax+by=c$ where $a,b$ and$c$ are integers to be found.

First find the gradient of the line using 'rise over run'

$m = \frac{- 7 - 5}{6 - ( - 2 )} = \frac{- 12}{8} = - \frac{3}{2}$

*Substitute the gradient and the coordinates of one of the points into *$y - y_{1} = m ( x - x_{1} )$

$\begin{array}{rcl} y - 5 & = - \frac{3}{2} ( x - ( - 2 ) ) \\ y - 5 & = - \frac{3}{2} ( x + 2 ) \\ y - 5 & = - \frac{3}{2} x - 3 \end{array}$

Multiply both sides of the equation by 2 to get rid of the fraction

$\begin{gathered} 2 ( y - 5 ) & = 2 ( - \frac{3}{2} x - 3 ) \\ 2 y - 10 & = - 3 x - 6 \end{gathered}$

Rearrange into the required form

$3x+2y=4$

Parallel Lines

Spec point — 8E

How are the equations of parallel lines connected?

• Parallel lines are always equidistant meaning they never intersect

• Parallel lines have the same gradient

  • If the gradient of line $l_{1}$ is $m_{1}$ and the gradient of line $l_{2}$ is $m_{2}$ then

    • If $m_{1}=m_{2}$ then $l_{1}$ and $l_{2}$ are parallel
    • If $l_{1}$ and $l_{2}$ are parallel, then $m_{1}=m_{2}$

• To determine if two lines are parallel:

  • Rearrange into the form $y=mx+c$
  • Compare the gradients (i.e. the coefficients of $x$)
  • If they are equal then the lines are parallel

Worked Example The line $l$ passes through the point $(4,-1)$ and is parallel to the line with equation $2x-5y=3$ .

Find the equation of $l$ , giving your answer in the form $y=mx+c$.

First find the gradient of the given line from its equation

$\begin{gathered} 2 x - 5 y & = 3 \\ 5 y & = 2 x - 3 \\ y = \frac{2}{5} x - \frac{3}{5} \end{gathered}$

So the gradient of line $l$ is $\frac{2}{5}$

Insert that gradient and the coordinates of the point into $y - y_{1} = m ( x - x_{1} )$

$\begin{gathered} y - ( - 1 ) & = \frac{2}{5} ( x - 4 ) \\ y + 1 & = \frac{2}{5} x - \frac{8}{5} \end{gathered}$

Rearrange into the form required

$y=\frac{2}{5}x-\frac{13}{5}$

Perpendicular Lines

Spec point — 8E

How are the equations of perpendicular lines connected?

• Perpendicular lines intersect at right angles

• The gradients of two perpendicular lines are negative reciprocals

  • This means the product of their gradients is equal to $-1$

    • e.g. $\frac{1}{2}$ and $-2$

  • If the gradient of line $l_{1}$ is $m_{1}$ and the gradient of line $l_{2}$ is $m_{2}$ then...

    • If $m_{1}\times m_{2}=-1$ then $l_{1}$ and $l_{2}$ are perpendicular
    • If $l_{1}$ and $l_{2}$ are perpendicular, then $m_{1}\times m_{2}=-1$

• To determine if two lines are perpendicular:

  • Rearrange into the form $y=mx+c$
  • Compare the gradients (i.e. the coefficients of$x$)
  • If their product is $-1$ then the lines are perpendicular

• Be careful with horizontal and vertical lines

  • $x=p$ and $y=q$ are perpendicular where p and q are constants

Worked Example The line$l_{1}$ is given by the equation $3x-5y=7$.

The line$l_{2}$ is given by the equation $y=\frac{1}{4}-\frac{5}{3}x$ .

Determine whether $l_{1}$ and $l_{2}$ are perpendicular. Give a reason for your answer.

Rearrange the $l_{1}$ equation into $y=mx+c$ form

$\begin{gathered} 3 x - 5 y & = 7 \\ 5 y & = 3 x - 7 \\ y = \frac{3}{5} x - \frac{7}{5} \end{gathered}$

So the gradient of $l_{1}$ is $\frac{3}{5}$

The gradient of $l_{2}$ is $-\frac{5}{3}$ (don't be fooled by the order of the terms in the equation!) Find the product of the two gradients

$\frac{3}{5} \times ( - \frac{5}{3} ) = - \frac{15}{15} = - 1$

The product of the gradients is equal to $-1$, so the lines are perpendicular

Final answer: The product of the gradients of the two lines is equal to$-1$, so$l_{1}$and$l_{2}$are perpendicular