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

analytic geometrygeometry

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?

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 $$\left<\vec{AP},\frac{\vec{AB}}{\|\vec{AB}\|}\right>\frac{\vec{AB}}{\|\vec{AB}\|}.$$
PS : Given your picture, to simplify calculation, you should replace my $$A$$ by your $$B$$.