# Question #85982

Nov 17, 2016

$\left(- \frac{36}{25} , - \frac{23}{25}\right)$

#### Explanation:

Two methods of solving:

Method 1: Algebra

Let $\left({x}_{0} , {y}_{0}\right)$ be the point on $- 4 x + 3 y - 3 = 0$ which is closest to $\left(4 , - 5\right)$.

The line $- 4 x + 3 y - 3 = 0$ can be rewritten as $y = \frac{4}{3} x + 1$. The shortest line segment between a line and a point not on that line will be perpendicular to the line, meaning the line passing through $\left(4 , - 5\right)$ and $\left({x}_{0} , {y}_{0}\right)$ will be perpendicular to $y = \frac{4}{3} x + 1$.

If a line has slope $m$, then a perpendicular line to that line will have slope $- \frac{1}{m}$. Thus the line passing through $\left(4 , - 5\right)$ and $\left({x}_{0} , {y}_{0}\right)$ has slope $- \frac{1}{\frac{4}{3}} = - \frac{3}{4}$. Using the point-slope form a line, that gives it as

$y - \left(- 5\right) = - \frac{3}{4} \left(x - 4\right)$

$\implies y = - \frac{3}{4} x - 2$

As $\left({x}_{0} , {y}_{0}\right)$ lies on both that line and the original line, we have

$\left\{\begin{matrix}{y}_{0} = - \frac{3}{4} {x}_{0} - 2 \\ {y}_{0} = \frac{4}{3} {x}_{0} + 1\end{matrix}\right.$

$\implies - \frac{3}{4} {x}_{0} - 2 = \frac{4}{3} {x}_{0} + 1$

$\implies \frac{25}{12} {x}_{0} = - 3$

$\implies {x}_{0} = - \frac{36}{25}$

$\implies {y}_{0} = \frac{4}{3} \left(- \frac{36}{25}\right) + 1 = - \frac{23}{25}$

$\therefore \left({x}_{0} , {y}_{0}\right) = \left(- \frac{36}{25} , - \frac{23}{25}\right)$

Method 2: Calculus

As above, we can rewrite the line as $y = \frac{4}{3} x + 1$, meaning every point on the line is of the form $\left(x , \frac{4}{3} x + 1\right)$. Using the distance formula, the distance between a point on the line and $\left(4 , - 5\right)$ is

$\sqrt{{\left(4 - x\right)}^{2} + {\left(- 5 - \left(\frac{4}{3} x + 1\right)\right)}^{2}}$

$= \sqrt{{\left(4 - x\right)}^{2} + {\left(- \frac{4}{3} x - 6\right)}^{2}}$
$= \sqrt{\frac{25}{9} {x}^{2} + 8 x + 52}$

Our goal, then, is to find the $x$ which minimizes the above. As a distance is always positive, it will be minimal at the same point as its square is minimal, allowing us to square it to simplify calculations. This leaves us with the goal of minimizing

$\frac{25}{9} {x}^{2} + 8 x + 52$

To find its extrema, we first locate its critical point(s), that is, any point at which its derivative is $0$ or nonexistent. Taking its derivative, we get

$\frac{d}{\mathrm{dx}} \frac{25}{9} {x}^{2} + 8 x + 52 = \frac{50}{9} x + 8$

Setting it equal to $0$ and solving:

$\frac{52}{9} x + 8 = 0$

$\implies x = - 8 \cdot \frac{9}{50} = - \frac{36}{25}$

As this is its only critical point, any extremum must occur there. We could also verify that it is a minimum by using the second derivative test, if desired.

We now have the $x$ coordinate of the point in question. Plugging that into the equation of the line, we get

$y = \frac{4}{3} \left(- \frac{36}{25}\right) + 1 = - \frac{23}{25}$

As in the first solution, we arrive at the point $\left(- \frac{36}{25} , - \frac{23}{25}\right)$