Linear Equations II

Hong Kong

Stage 1 - Stage 3

Lesson

Lines that meet at right angles ($90^\circ$90°) are called perpendicular lines.

Play with this applet creating pairs of perpendicular lines.

Fill in this table as you go.

What do you notice about the product of the gradients of lines $1$1 and $2$2? (The pair of perpendicular lines)

You will have discovered the perpendicular lines have gradients whose product is equal to $-1$−1.

We say that $m_1$`m`1 is the **negative reciprocal** of $m_2$`m`2.

Negative reciprocal is a complex sounding term, but it basically means they have opposite signs and they are reciprocal of each other.

Here are some examples:

$2$2 and $\frac{-1}{2}$−12

$\frac{3}{4}$34 and $\frac{-4}{3}$−43

$-10$−10 and $\frac{1}{10}$110

**To test if lines are perpendicular simply multiply the gradients together. If the result is $-1$−1 then the lines are perpendicular. **

Find the equation of a line that is perpendicular to $y=-\frac{3x}{4}+7$`y`=−3`x`4+7, and goes through the point $\left(0,6\right)$(0,6).

Given the points $A$`A`$\left(6,10\right)$(6,10), $B$`B`$\left(-9,5\right)$(−9,5), $C$`C`$\left(-3,-2\right)$(−3,−2) and $D$`D`$\left(-6,7\right)$(−6,7), let $m$`m` and $n$`n` be the gradients of intervals $AB$`A``B` and $CD$`C``D` respectively.

Prove that interval $AB$`A``B` is perpendicular to the interval $CD$`C``D` by showing that $mn=-1$`m``n`=−1.

Consider the line $5x-3y-9=0$5`x`−3`y`−9=0.

Find the $y$

`y`-intercept of the line.Find the equation of the line that is perpendicular to the given line and has the same $y$

`y`-intercept. Express in general form.