Distance Question - Shortest Possible Distance Question: A straight line passing through the point $(8,27)$ intersects the positive x-axis at the point $P$ and the positive y-axis at the point $Q$. What is the shortest possible distance between $P$ and $Q$.
So from this
I concluded that $P = (x, 0)$ and $Q = (0, y)$
I made $R = (8, 27)$
I think I need to find the distance from $R$ to $P$ and $R$ to $Q$ and add them together to get the total, however, I don't understand how that would be the shortest possible distance. Also, when I tried doing this, I ended up with 2 variables. 
Update: Everybody seems to be getting different results.... If it helps I'm in first year calculus and all we were given for this unit was the distance equation and how to find an equation from a tangent line and a point
 A: If the three points $P,Q,R$ are on a line, then for a suitable $t\in \mathbb R$ we have
$$ R = (1-t) P + t Q. $$
So you have 2 equations for 3 unknowns. Now, for example, solve $x$ and $y$ for $t$ and minimize 
$$ x^2(t) + y^2(t). $$
A different approach:
Say $m$ is the slope of the line going through $P,Q,R$. Then, we have
\begin{align*}
0 &= 27 + m(x - 8) \\
y &= 27 + m(0 - 8)
\end{align*}
Now, again, solve for $x$ and $y$ and minimize
$$ x^2(m) + y^2(m). $$
A: My idea is to set the angle between $PQ$ and the $x$-axis as $\theta$. Hence, the length of $PQ$ is $$|PQ| = \frac{8}{\cos \theta}+\frac{27}{\sin\theta} :=f(\theta)\implies f'(\theta)=8\cdot\frac{\sin\theta}{\cos^2\theta}-27\cdot\frac{\cos\theta}{\sin^2\theta}$$
Solving the above equation for $f'(\theta )=0$, we know the derivative is $0$ when $\tan\theta = \frac{3}{2}$, and this gives the shortest distance because the boundary cases with $\theta\to 0^\circ$ and $\theta\to 90^\circ$ results in $|PQ|\to\infty$. Therefore, the shortest distance occurs when $$\cos\theta = \frac{2}{\sqrt{13}},\ \sin\theta = \frac{3}{\sqrt{13}} \implies |PQ|=13\sqrt{13}$$
A: Your line should be of the form $y=m(x-8)+27$
Rewriting, we have $y=mx-8m+27$ or $\boxed{y=mx+(27-8m)}$ for y-intercept
Let's rewrite again $y-27=m(x-8)$ to $\frac{y}{m}-\frac{27}{m}+8=x$.
We have $\boxed{x=\frac{y}{m}+(8-\frac{27}{m})}$ for x-intercept.
Our y-intercept is $(0,27-8m)$ and our x-intercept is $(8-\frac{27}{m},0)$
Pythagorean theorem gives us the distance as $\sqrt{(27-8m)^2+(8-\frac{27}{m})^2}$.
Differentiate with respect to $m$ to get some nasty derivative, which is equal to zero at $m=\frac{3}{2}$. 
The distance would be equal to $5\sqrt{13}$ units.
A: Oh dear.  Let's do this using only properties of derivatives, the usual point-slope equation of a line, and some algebra.
A line that passes through a point $R = (8,27)$ has the equation $$y - 27 = m(x-8),$$ for some slope $m$.  Thus the only variable that determines the points $P$ and $Q$ is the slope $m$.  The $x$-intercept is the value of $x$ when $y = 0$, so this is a function of $m$: $$P(m) = -\frac{27}{m} + 8.$$  Similarly, the $y$-intercept is the value of $y$ when $x = 0$:  $$Q(m) = -8m + 27.$$  It follows that the distance $PQ$ is a function of $m$ given by $$|PQ(m)| = \sqrt{P(m)^2 + Q(m)^2} = \sqrt{\frac{(27-8m)^2 (1+m^2)}{m^2}}.$$  Because distances are nonnegative, the distance is minimized when the squared distance is minimized; i.e., we can find the $m$ that minimizes $|PQ(m)|^2$ rather than $|PQ(m)|$.  To this end, we simply calculate the critical points:  $$0 = \frac{d}{dm}\left[|PQ(m)|^2 \right] = \frac{2}{m^3}(8m-27)(8m^3+27).$$  So $$m \in \left\{\frac{27}{8}, -\frac{3}{2}\right\}.$$  In the first case, the equation of the line passes through $(0,0)$:  this is not allowed, since we are told the $x$- and $y$-intercepts are positive.  Thus the only choice is $m = -3/2$, leading to the line $$y = -\frac{3}{2} x + 39,$$ and the minimum distance is $$|PQ(m)| = 13 \sqrt{13}.$$

