Find the point on the line $x + 4y − 7 = 0$ which is closest to the point $(-2, -2)$ Find the point on the line
$$x + 4y − 7 = 0$$
which is closest to the point $(-2, -2)$
First I used the distance formula and found I need to minimize
$$(x+2)^2 + \Big(\frac{7}{4}-\frac{x}{4}+2\Big)^2$$
Taking the derivative gave me
$$(2x+4)-\frac{7}{8}+\frac{x}{8}-1$$
Simplifying gave me $x=-1$
Is this right?
 A: People are giving you calculus answers where you have to minimize a distance function, but because this is a straight line, such machinery is not needed. 
Suppose you have a point $P$ and a line $l$ not passing through $P$. Draw a line $l'$ perpendicular to $l$ and passing through $P$, and label the intersection of $l$ and $l'$ as $Q$. Then, if $Q'$ is any other point on the line $l$, you can use properties of triangles to show $PQ<PQ'$. But this is the same as saying that $Q$ is the closest point to $P$ on $l$. So we have the ancient theorem that to find the point on a straight line that is closest to a given point, drop a perpendicular.
In particular, your line $l$ in slope-intercept form is 
$$y=-\frac{1}{4} x + \frac{7}{4}$$
and has slope $-\frac{1}{4}$. The perpendicular $l'$ will have slope $4$ and we want it to pass though $P=(-2,-2)$. Thus $l'$ has the equation
$$(y+2)=4(x+2).$$
Solving the system consisting of the two displayed equations will find the intersection of $l$ and $l'$, your point $Q$.  That point will be closest to the given point while still lying on the given line.
It is also quite easy to find the point on a given circle that is closest to a given point not on the circle. But for more complicated curves, you will need to formulate a distance function and minimize it, and that will require calculus.
A: Rachel, what have you tried? You should post where you failed so people are more inclined to help you. You want to minimize the distance function $s^2 = (x-x_0)^2 + (y-y_0)^2$ where $(x_0,y_0)=(-2,2)$ in this case. Once you have this $s^2$ equation, you can plug in your equation for $x$ or $y$ (from the equation for a line) and minimize your $s$ function by taking the derivative and finding when it is zero. This x or y-value can then be used in your $s$ equation.
A: Take an arbitrary point of your line, say $P_0 =(x, \frac{7}{4} - \frac{x}{4})$. We want to minimize the distance between $P_0$ and $P_1 = (-2,-2)$.
$$ \therefore d^2(P_0, P_1) = (x+2)^2 + (\frac{7}{4} - \frac{x}{4} + 2)^2 = f(x)$$
Now, apply what you have learned from calculus. For instance, you must find the critical points. That is, you want to find what is $\{ x : f'(x) = 0 \} $.  And you know what to do next.
