Finding the distance of the line to apoint

Question

Find the distance from $3x-4y-10=0$ to the point $(2,0)$

my answer here is $ \dfrac{-4}{2}$ or $-2$ by substituting the given by the use of the formula but Im just wondering if there's a negative distance because it can be possibly rewritten to $ \dfrac{-4}{-2}$ or $2$. Because the square root of $4$ can be positive or negative $2$. What is the correct answer and explanation here?

Answer 1

I think you have the right formula, check Point-Line Distance (2D). The distance from $(x_0,y_0)$ to the line with equation $ax+by+c=0$ is given by $$d = \frac{\left| ax_0+by_0+c \right|}{\sqrt{a^2+b^2}}$$ Check your calculations and watch out for the absolute value in the numerator. The square root in the denominator is, by definition, (the) positive (square root).

In your case $(x_0,y_0) = (2,0)$ and $a=3$, $b=-4$, $c=-10$.

Answer 2

$$ d= \frac{|Ax + By + C|}{\sqrt{ A^2 + B^2}}=\dfrac{\left| 3\left( 2\right) -4\left( 0\right) -10\right| }{\sqrt{\left( 3^{2}\right) +\left( -4^{2}\right) }}=\dfrac{4}{5} $$