Finding the Interception Points of a Line

# Finding the Interception Points of a Line

## Calculating Coordinates of x-Intercepts

Suppose that we have a line and we want to find the coordinates of the $x$-intercepts. Recall that coordinates of an $x$-intercept always have their $y$-coordinate at 0. Imagine you're in the air. We could analogously imagine a solution to occur if you touch the ground. You'll only touch the ground when your height from the ground is at 0 though. The same concept holds for x-intercepts.

If we take an equation in the form $y = mx + b$ or $ax + by + c = 0$ and set $y$ as 0 and solve for $x$, we will find the $x$-coordinate that corresponding to the $y$-coordinate of 0, or more appropriately, we will have the coordinates of our $x$-intercept.

For example, let's look at the line $y = 2x + 4$. If we set $y$ to be 0, then all we have to do is solve for a 1-variable linear equation:

(1)
\begin{align} y = 2x + 4 \\ 0 = 2x + 4 \\ -4 = 2x \\ -2 = x \end{align}

Therefore, we have an $x$-intercept at the coordinates $(-2, 0)$.

We can apply this same rule to lines in general form. For example, consider the line $x -3y + 7 = 0$. Once again, we will set $y$ to be 0, and thus:

(2)
\begin{align} x - 3y + 7 = 0 \\ x - 3(0) + 7 = 0 \\ x + 7 = 0 \\ x = -7 \end{align}

Therefore we have an $x$-intercept at $(-7, 0)$.

We should note that finding $x$-intercepts in general form is easy. Suppose that $ax + by + c$ is any line, and set $y$ to be 0:

(3)
\begin{align} ax + by + c = 0 \\ ax + b(0) + c = 0 \\ ax + c = 0 \\ ax = -c \\ x = \frac{-c}{a} \end{align}

Therefore, we note that a line in general form has an $x$-intercept at $(\frac{-c}{a}, 0)$.

## Calculating Coordinates of y-intercepts

If we have a line in slope-intercept form $y = mx + b$, we can easily obtain the $y$-intercept from the line to have the coordinates $(0, b)$. For example, the line $y = x + 4$ has a $y$-intercept at $(0, 4)$.

If we have a line in general form, we could always convert it to slope-intercept form to obtain the coordinates of the $y$-intercept. However, recall that $y$-intercepts always have their $x$-coordinate as 0, so similarly to how we found $x$-intercepts, we can always set $x$ to be 0 and solve for $y$.

For example, take the equation $2x + 3y + 4 = 0$ and set $x$ to be 0, then solve the 1-variable linear equation:

(4)
\begin{align} 2x + 3y + 4 = 0 \\ 2(0) + 3y + 4 = 0 \\ 3y + 4 = 0 \\ 3y = -4 \\ y = \frac{-4}{3} \end{align}

Therefore our line has a $y$-intercept at $(0, \frac{-4}{3})$. If we had converted this line to slope-intercept form, we would have seen the same results.

In general, we note that if we have a line in $ax + by + c$ form and set $x$ to be 0:

(5)
\begin{align} ax + by + c = 0 \\ a(0) + by + c = 0 \\ by + c = 0 \\ by = -c \\ y = \frac{-c}{b} \end{align}

We thus have that our $y$-intercept has coordinates $(0, \frac{-c}{b})$.

It is important to note the similarity and differences between calculating coordinates for x-intercepts and y-intercepts.

## Example Questions

• 1. Calculate the coordinates of the x-intercepts and y-intercepts for the line $y = 5x - 3$.
• 2. Calculate the coordinates of the x-intercepts and y-intercepts for the line $y = 4$.
• 3. Calculate the coordinates of the x-intercepts and y-intercepts for the line $x = -3$.
• 4. Calculate the coordinates of the x-intercepts and y-intercepts for the line $2x - 3y + 7 = 0$.