Math help?

Question

I need help with this problem.

A line with slope m passes through the point (0,4) and has the equation y=mx+b.
a) write the distance d between the line and the point (3,1) as a function of m.

its in the derivative chapter. but i don't understand how to go about solving it and if i even need to find any derivatives on it.
Thanks.

Answer 1

OK... y = mx + b passes through (0,4). When x=0, y=4, so b must equal 4.

The shortest distance will be on a line that's at right angles to your y=mx+4 line. Therefore, you need to find a line that's at right angles to y=mx+4, and passes through the point (3,1).

This will be the line y = -(1/m)x + c, where the values of m and c work for the point (3,1).

Once you have that line, find the intersection between that line and your original y=mx+4 line.

mx+4 = -(1/m)x + c
m^2x + 4m = -x + mc
m^2x +x +4m = mc - 4m
x(m^2 + 1) = mc - 4m
x = (mc - 4m) / (m^2 + 1)

That's the x-coordinate of the closest point on y=mx+b.
Given the x-coordinate, you can find the y-coordinate.

To find the distance, find the difference between the x-coordinate and 3, and the difference between the y-coordinate and 1, and use Pythagoras. I'll let you work out the details :-)

Answer 2

This is not a derivative problem - this is finding a distance from a point to a line. Use the point-line distance formula.