Question

what is the distance between the y-intercept and the x-intercept of the line whose equation is y=3x+12

Answer 1

The distance between the y-intercept and the x-intercept of the line is 12.64911 units or $\sqrt{160}$ units

Solution:

Given equation is y = 3x + 12

We have to find the distance between y-intercept and the x-intercept of the line whose equation is y = 3x + 12

Let us first find the x - intercept and y - intercept

The x-intercept is the point at which the line crosses the x-axis. At this point, the y-coordinate is zero.

The y-intercept is the point at which the line crosses the y-axis. At this point, the x-coordinate is zero

Finding x - intercept:

To find the x-intercept of a given linear equation, plug in 0 for 'y' and solve for 'x'.

Substitute y = 0 in given equation

y = 3x + 12

0 = 3x + 12

3x = -12

x = -4

Thus the x - intercept is (-4, 0)

Finding y - intercept:

To find the y-intercept, plug 0 in for 'x' and solve for 'y'

Substitute x = 0 in given equation

y = 3x + 12

y = 3(0) + 12

y = 12

Thus the y - intercept is (0, 12)

Now let us find the distance between x intercept and y intercept

Distance between two points $P(x_1, y_1) \text{ and } Q(x_2, y_2)$ is given by:

$d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$

Here the distance between (-4, 0) and (0, 12) is given as:

$\begin{aligned}&d=\sqrt{(0-(-4))^{2}+(12-0)^{2}}\\\\&d=\sqrt{4^{2}+12^{2}}\\\\&d=\sqrt{16+144}=\sqrt{160}\\\\&d=12.64911\end{aligned}$

Thus the distance between the y-intercept and the x-intercept of the line is 12.64911 units or $\sqrt{160}$ units