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 $B$.