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# How do you graph $y=-3x-3$ using rise over run?

Last updated date: 20th Jun 2024
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Hint: The ratio rise over run is equal to the slope of the line. The equation of a line is given as $y=mx+c$ , where m is the slope and c is y intercept. Compare the two equations and find the value of the ratio rise over run and find the x intercept of the line.

Complete step-by-step solution:
The given equation is a linear equation in two variables. When we plot a graph of a linear equation in two variables on a Cartesian coordinate plane, we will get a straight line.
Therefore, the given equation in x and y variables represents a straight line graphically.
Let us now see how to plot a straight line with its equation using the rise over run method.
To plot any line we need the two intercepts of the line. This can be done using rise over run.
The ratio rise over run is equal to the slope of the line. The equation of a line is given as $y=mx+c$ , where m is the slope and c is y intercept.
Comparing the general equation with $y=-3x-3$ we get that $m=-3$ and $c=-3$
We know that $\dfrac{rise}{run}=m$ and $m=-3$
$\Rightarrow \dfrac{rise}{run}=-3$ … (i)
We know that the y intercept of the line is equal to -3 units. For a rise equal to the y intercept, there is a run equal to the x intercept.
Therefore, $rise=c=-3$ and $run=a$ , where a is x intercept.
Substitute this values in (3)
$\Rightarrow \dfrac{-3}{a}=-3$
$\Rightarrow a=1$
This means that the x intercept of the line is 1.
Since we know the x and y intercepts of the line, we can plot the line on a Cartesian plane as shown below.

Note: This not the only way in which we can plot a line with its equation given. We can also use the slope and find the angle that the line makes with the positive x-axis. With this angle and the y intercept we can plot the line. We can also plot the straight line by knowing the x and y intercepts of the line. The equation of the line in the terms of x and y intercepts is given as $\dfrac{x}{a}+\dfrac{y}{b}=1$ , where a and b are the x and y intercepts respectively.