Question

How do you find the $x$ and $y$ intercept given $2x - 3y - 12 = 0$?

Answer 1

Hint: Here we will find the $x$ and $y$ intercept of a given straight line equation. By considering first $x$ intercept and then $y$ intercept. On doing some simplification we get the required answer.

Complete step-by-step solution:
The $x$ intercepts of a graph $y = f(x)$ are the $x$ coordinates of the points where the graph hits the $x$ axis and the $y$ intercept is the $y$ coordinate of the point where the graph hits the $y$ axis.
If we know the $x$ intercept, we should have coordinates ($x$ value,$0$). If we know the $y$ intercept, we should have another coordinate ($y$ value,$0$).
Plot the points on the graph starting right or left with the $x$ value and then up or down with the $y$ value, finding where they intersect and placing a point. We should have two points when we are done. Connect them using a ruler or straight edge and you have graphed a line.
Now for the given equation $2x - 3y - 12 = 0$
The $x$ intercept is where the graph crosses the $x$ axis, which is where $y = 0$.
Similarly, the $y$ intercept is where the graph crosses the $y$ axis, or where $x = 0$.
To find the intercept points, plug in $0$ for either $x$ or $y$ and solve for the other.
Let’s start with the $x$ intercept.
$ \Rightarrow 2x - 3\left( 0 \right) - 12 = 0$
Add $12$ to both sides. We get,
$ \Rightarrow 2x = 12$
Divide both sides by $2$,
$ \Rightarrow x = 6$
So the ordered pair for the $x$ intercept is $\left( {6,0} \right)$
Now do the same with the $y$ intercept.
Putting $x = 0$, we get,
$ \Rightarrow 2\left( 0 \right) - 3y - 12 = 0$
Again, Add $12$ to both sides. We get,
$ \Rightarrow - 3y = 12$
This time divide by $ - 3$ we get,
$ \Rightarrow y = - 4$
So the ordered pair for the $y$ intercept is $\left( {0, - 4} \right)$

Note: It is possible for a line to have an infinite number of intercepts with the $x$ or $y$ axis. The line $x = 0$ has an infinite number of intercepts with the $y$ axis. The line $y = 0$ has an infinite number of intercepts with the $x$ axis. Any line of the form $y = mx + b$ where $m \ne 0$ has exactly one $y$ intercept. If $b = 0$ then both the $x$ and $y$ intercepts are at the origin $\left( {0,0} \right)$.