Find an equation perpendicular to y=(-3/5)x+1 and passing through (5,-4)

An equation perpendicular to this would have slope 5/3. Use the point-slope form of the line and multiply out to find the final answer:

(y+4)/(x-5) = 5/3

3(y+4) = 5(x-5)

3y+12 = 5x-25

3y=5x-25-12 = 5x-37

y = 5/3x-37/3

Check:

y = 5/3(5)-37/3=25/3-37/3=-12/3=-4

So the equation does pass through the point (5,-4).

y = (-3/5)x + 1

When finding an equation that perpendicular, you're looking for an equation with the negative reciprical slope. (e.g. 5/3 is the negative reciprical of -3/5)

y = (5/3)x + b <--- Now look for b
-4 = 25/3 + b <--- plug in y and x value to solve for b
b = -12/3 - 25/3
b = -9

y = (5/3)x - 9

Didn't I just answer this? :-)

OK, this is how you do it. The perpendicular line has the opposite slope, 5/3. It is opposite both in sign and in that it is the inverse of the slope of your equation. Once you know the slope, you fill in the point you know, and solve for b:

-4 = (5/3) 5 + b

b=-37/3

So

y = (5/3)x-37/3, y=(-3/5)x+1

WolframAlpha graphs your lines in the first reference

show2d(ploteq2d(y=(5/3)x-37/3,{x, -15.891, 30.945},{y, -15.001, 7.999}),ploteq2d(y=((-3)/5)x+1,{x, -15.891, 30.945},{y, -15.001, 7.999}),{Proportional, true},{x, -15.891, 30.945},{y, -15.001, 7.999})

That command in Microsoft Math graphs your equations using proportional coordinates and you can clearly see that they are perpendicular.

It also shows the line going through your requested point.