# Find the coordinates of the point on the line y = 3x + 1 closest to (4, 0).

All we need to do to solve this question is to find another line passing through the point (4,0) and perpendicular (orthogonal line) to y = 3x + 1 which will give us the shortest distance between the line and that point.

First of all, to know the inclination of the orthogonal line to y = 3x + 1 given by:

$m&space;=&space;-&space;\frac{1}{m_{1}}$

we will need to find the inclination of the line y = 3x + 1, which can be evaluated by:

$m_{1}&space;=&space;3&space;\,\,\,\,\,\,\,&space;Since\,\,&space;y&space;-&space;y_{0}&space;=&space;m_{1}(x&space;-&space;x_{0})&space;\,\,\,\,&space;where\,\,&space;y_{0}&space;=&space;1&space;\,\,&space;and&space;\,\,&space;x_{0}&space;=&space;0$

Thus:

$m&space;=&space;-\frac{1}{3}$

We already know that the point (4,0) belongs to the set of points of the line whose inclination is -1/3. Using the formula to get the Equation of a line we have:

$y&space;-&space;y_{0}&space;=&space;m(x&space;-&space;x_{0})$

$y&space;-&space;0&space;=&space;-\frac{1}{3}&space;(x&space;-&space;4)$

$y&space;=&space;-\frac{x}{3}&space;+&space;\frac{4}{3}$

Now, we will need the point of intersection of those two lines we have now. This point is the closest point to the point (4,0). To accomplish this, all we have to do is:

$-\frac{x}{3}&space;+&space;\frac{4}{3}&space;=&space;3x&space;+&space;1$

$-x&space;+&space;4&space;=&space;9x&space;+&space;3$

$-3&space;+&space;4&space;=&space;9x&space;+&space;x$

$10x&space;=&space;1$

$x&space;=&space;\frac{1}{10}$

Using this result in any one of the line equations we already have (I am going to use the first given line equation):

$y&space;=&space;3x&space;+&space;1&space;\Rightarrow&space;y&space;=&space;3&space;\left&space;(&space;\frac{1}{10}&space;\right)&space;+1&space;=&space;\frac{3&space;+&space;10}{10}&space;=&space;\frac{13}{10}$

Thus, the point (1/10, 13/10) is the closest point in both lines to the point (4,0).

See the graph below:

$\blacksquare$