Linear Graphs

Equations of a Straight Linevideo

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 between these two points is calculated by

$m=\frac{y_2-y_1}{x_2-x_1}$ 

• 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 the gradient-intercept form

  • It clearly shows the gradient m and the y-intercept (0, c)

• $y-y_1=m\left(x-x_1\right)$

• This is the point-gradient form

  • It clearly shows the gradient m and a point on the line (x₁, y₁)

• $ax+by+d=0$

  • This is the general form

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

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

  • then rearrange into whatever form is required

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

Midpoint of a Straight Line

Midpoint of a line segment

How do I find the midpoint of a line segment?

• The midpoint of a line will be the same distance from both endpoints

• You can think of a midpoint as being the average (mean) of two coordinates

• The midpoint of $\left(x_1, y_1\right)$ and $\left(x_2, y_2\right)$ is

$\left(\frac{x_1+x_2}{2} , \frac{y_1+y_2}{2}\right)$

• M is often used as the midpoint between two points (x₁, y₁) (x₂, y₂)

• It is the average of both the x and y coordinates

Length of a Straight Line

Length of a line segment

How do I calculate the length of a line segment?

• The distance between two points with coordinates $\left(x_1 , y_1\right)$ and $\left(x_2 , y_2\right)$ can be found using the formula

$d=\sqrt{{\left(x_1-x_2\right)}^2+{\left(y_1-y_2\right)}^2}$

• This formula is really just Pythagoras’ Theorem  $a^2=b^2+c^2$, applied to the difference in the $x$-coordinates and the difference in the $y$-coordinates;

• You may be asked to find the length of a diagonal in 3D space

  • This can be answered using 3D Pythagoras