What is the equation of a line that passes through the point (10, 5) and is perpendicular to the line whose equation is y=5/4x−2?

The slope of the given line is 5/4.  The slope of a line perpendicular to the given line is the negative reciprocal of 5/4, or -4/5.

Using the point-slope form, 

y - 5 = (-4/5)(x - 10).

This is the desired new line.

Note that mult. both sides by 5 removes the fractions:

5y - 25 = -4(x - 10), or     5y - 25 = -4x + 40.

Solving this for y:   5y = 25 - 4x + 40, or 5y = -4x + 65

Then y = (-4/5)x + 13 (answer)