Find Distance between a Point $(-4,2)$ and a Line having equation $3x−y−10=0$

My attempt:

I used a graphing calculator to graph the equation $3x−y−10=0$ and the point $(-4,2)$. I found the x-intercept and y-intercept of the equation as $\dfrac{10}{3}$, and $-10$ respectively.

I used the Midpoint Formula to get the midpoint of $AB$;

$$

\left(x_{m}, y_{m}\right)=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)

$$

$$= \left(\dfrac{\dfrac{10}{3} + 0}{2}, \; \dfrac{0 + -10}{2}\right) = \left(\dfrac{5}{3}, \; -5\right)$$

Now, Can I use the distance formula $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ to find the distance between the point and the midpoint of the line? By distance formula, $d=9$. Is my approach correct?

But the solution given in my textbook uses some other distance formula, $d=\left| \dfrac{ax_{1}+by_{1}+c}{\sqrt{a^{2}+b^{2}}}\right|$. Can you please explain to me how we got this formula and how to solve this problem with this formula?

## Best Answer

Forget about formula. In the following draw, the distance between $P$ and the line passing through $A$ and $B$ is $\|\vec{PH}\|$. Now, $$\vec{PH}=\vec{AH}-\vec{AP}.$$

You know $\vec{AP}$. Now, $\vec{AH}$ is the orthogonal projection of $\vec{AP}$ on $\vec{AB}$, and this is $$\left<\vec{AP},\frac{\vec{AB}}{\|\vec{AB}\|}\right>\frac{\vec{AB}}{\|\vec{AB}\|}.$$

You have all element to conclude.

PS : Given your picture, to simplify calculation, you should replace my $A$ by your $B$.