Equation of a Line: Standard Form
There are a number of ways of stating an equation for a straight line. Through the next few chapters we will be looking at them, and here is the first one:
A line equation is in standard form if its equation looks like
$ax+by=c$ax+by=c
where $a$a, $b$b, and $c$c are all integers and the value of $a$a is non-negative, that is, $a>0$a>0.
question 1
Express the following equations in standard form:
a. $y=4x-5$y=4x−5
Think: We need to rearrange the equation so that $x$x and $y$y are on the same side, and make sure that the coefficient of $x$x is positive. Since this coefficient is already positive, we should bring the $y$y over to it.
Do:
| $y$y | $=$= | $4x-5$4x−5 |
| $0$0 | $=$= | $4x-5-y$4x−5−y |
| $5$5 | $=$= | $4x-y$4x−y |
| $4x-y$4x−y | $=$= | $5$5 |
b. $y=\frac{-4x}{5}-1$y=−4x5−1
Think: This time the coefficient of $x$x is negative, so we will bring the $x$x-term over to the $y$y-term. We also need to multiply by $5$5 to remove the fractions.
Do:
| $y$y | $=$= | $\frac{-4x}{5}-1$−4x5−1 |
| $y+\frac{4x}{5}$y+4x5 | $=$= | $-1$−1 |
| $5y+4x$5y+4x | $=$= | $-5$−5 |
question 2
The equation of a straight line is given by $4x-9y=-5$4x−9y=−5. Find the $y$y-intercept.
Think: At the $y$y-intercept we know $x=0$x=0, so we substitute that value of $x$x into the equation.
Do:
| $4\times0-9y$4×0−9y | $=$= | $-5$−5 |
| $-9y$−9y | $=$= | $-5$−5 |
| $y$y | $=$= | $\frac{-5}{-9}$−5−9 |
| $y$y | $=$= | $\frac{5}{9}$59 |
question 3
A line has gradient $\frac{14}{3}$143 and passes through the point $\left(-5,-10\right)$(−5,−10).
By substituting into the equation $y=mx+b$y=mx+b, find the value of $b$b for this line.
Write the equation of the line in standard form.