Question

find the distance from the line to the given point.
$y = \frac{1}{6}x + 6$; $(-6,5)$

units

Explanation:

Step1: Recall the distance formula

The distance $d$ from a point $(x_0,y_0)$ to a line $Ax + By+C = 0$ is given by $d=\frac{\vert Ax_0+By_0 + C\vert}{\sqrt{A^2 + B^2}}$. First, rewrite the line $y=\frac{1}{6}x + 6$ in the general form $Ax+By + C=0$.
$y=\frac{1}{6}x + 6$ can be rewritten as $x-6y+36 = 0$, where $A = 1$, $B=-6$, $C = 36$, and the point $(x_0,y_0)=(-6,5)$.

Step2: Substitute values into the formula

Substitute $x_0=-6$, $y_0 = 5$, $A = 1$, $B=-6$, and $C = 36$ into the distance formula $d=\frac{\vert Ax_0+By_0 + C\vert}{\sqrt{A^2 + B^2}}$.
We get $d=\frac{\vert1\times(-6)+(-6)\times5 + 36\vert}{\sqrt{1^2+(-6)^2}}$.
First, calculate the numerator: $\vert-6-30 + 36\vert=\vert-36 + 36\vert=0$.
The denominator is $\sqrt{1 + 36}=\sqrt{37}$.
So $d=\frac{\vert0\vert}{\sqrt{37}}=0$.

Answer:

$0$