Linear Equations II

Hong Kong

Stage 1 - Stage 3

Lesson

We know that the gradient of a line is a measure of its steepness or slope.
## Horizontal Lines

Straight lines on the Cartesian Plane can literally be in any direction and pass through any two points.

This means that straight lines can be:

Remember!

$Gradient=\frac{rise}{run}$`G``r``a``d``i``e``n``t`=`r``i``s``e``r``u``n`

On horizontal lines, the $y$`y` value is always the same for every point on the line. In other words, there is no rise- it's completely flat.

$A=\left(-4,4\right)$`A`=(−4,4)

$B=\left(2,4\right)$`B`=(2,4)

$C=\left(4,4\right)$`C`=(4,4)

All the $y$`y`-coordinates are the same. Every point on the line has a $y$`y` value equal to $4$4, regardless of the $x$`x`-value.

The equation of this line is $y=4$`y`=4.

Since gradient is calculated by $\frac{\text{rise }}{\text{run }}$rise run and there is no rise (ie. $\text{rise }=0$rise =0), the gradient of a horizontal line is always $0$0.

On vertical lines, the $x$`x` value is always the same for every point on the line.

Let's look at the coordinates for A,B and C on this line.

$A=\left(5,-4\right)$`A`=(5,−4)

$B=\left(5,-2\right)$`B`=(5,−2)

$C=\left(5,4\right)$`C`=(5,4)

All the $x$`x`-coordinates are the same, $x=5$`x`=5, regardless of the $y$`y` value.

The equation of this line is $x=5$`x`=5.

Vertical lines have no "run" (ie. $\text{run }=0$run =0). l If we substituted this into the $\frac{\text{rise }}{\text{run }}$rise run equation, we'd have a $0$0 as the denominator of the fraction. However, fractions with a denominator of $0$0 are undefined.

So, the gradient of vertical lines is *always* undefined.

Did you know?

Linear equations can be written in the form $y=mx+b$`y`=`m``x`+`b`, where $m$`m` is the gradient.

Notice how the equations of horizontal and vertical lines are *not* written in this form. Neither of them have a coefficient of $x$`x`.

What is the gradient of any line parallel to the $x$`x`-axis?

$A$`A` $\left(2,1\right)$(2,1), $B$`B` $\left(7,3\right)$(7,3) and $C$`C` $\left(7,-5\right)$(7,−5) are the vertices of a triangle.

Which side of the triangle is a vertical line?

$BC$

`B``C`A$AB$

`A``B`B$AC$

`A``C`CDetermine the area of the triangle using $A=\frac{1}{2}bh$

`A`=12`b``h`.