Linear Equations I

Hong Kong

Stage 4 - Stage 5

Lesson

In coordinate geometry, we are often asked to find the equation of a line. However, as you have learnt, there are lots of rules we can use to find equations. Let's go over them now.

To use the gradient-intercept rule, we need to be able to find the gradient and the $y$`y`-intercept of a line.

It's written in the form:

$y=mx+b$`y`=`m``x`+`b`

where $m$`m` is the gradient and $b$`b` is the $y$`y`-intercept

Write down the equation of line L1 in gradient-intercept form, given that its slope is $-3$−3 and its $y$`y`-intercept is $-2$−2.

Think: If we consider the gradient-intercept form of the line ($y=mx+b$`y`=`m``x`+`b`), we can find immediately find the equation if we know the gradient and the $y$`y`-intercept.

Do: Substitute $m=-3$`m`=−3 and $b=-2$`b`=−2 into the gradient-intercept form.

Solution: The equation of line L1 is $y=-3x-2$`y`=−3`x`−2

If we don't know the $y$`y`-intercept of the line but we do know a point on the line, we can use the point-gradient formula, which is:

$y-y_1=m\left(x-x_1\right)$`y`−`y`1=`m`(`x`−`x`1)

A line passes through the point $A$`A`$\left(-4,3\right)$(−4,3) and has a gradient of $-9$−9. Using the point-gradient formula, express the equation of the line in gradient intercept form.

We can also use two pairs of coordinates to find the equation of a line. All we have to do is substitute the coordinates into the two point formula:

$\frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1}$`y`−`y`1`x`−`x`1=`y`2−`y`1`x`2−`x`1

Use the two point formula to derive the equation of the line that passes through the points $A$`A`$\left(6,5\right)$(6,5) and $B$`B`$\left(-16,11\right)$(−16,11). Give your answer in general form.

Another way of writing the equation of a straight line is called general form. A straight line is in general form when it is written with all terms on one side of the equation and zero on the other, with integer coefficients. In particular, the coefficient of $x$`x` should be positive (the sign can be easily changed by multiplying the whole equation by $-1$−1).

That is, a straight line is in general form when it is of the form

$ax+by+c=0$`a``x`+`b``y`+`c`=0

for integers $a$`a`, $b$`b` and $c$`c` with $a>0$`a`>0.

Neil wanted to rewrite $y=\frac{7x}{5}+10$`y`=7`x`5+10 in general form. His steps are as follows:

Step $1$1: $y=\frac{7x+50}{5}$`y`=7`x`+505

Step $2$2: $5y=7x+50$5`y`=7`x`+50

Step $3$3: $7x+5y+50=0$7`x`+5`y`+50=0

In which step did Neil make a mistake?

Step $2$2

AStep $3$3

BStep $1$1

C