Question

A line Q has an equation of x+5y+5=0. Find the equation of the line that is "perpendicular" to Q an passes through (3,1). Find the equation of the line that is "parallel" to Q and passes to through (3,1). Find the equation of the line that passes through (3,1) that makes an "angle of 45 degrees clockwise" from the line that is perpendicular to line Q.

Answer 1

Line Q has slope -1/5; so do all parallel lines
Perpendicular lines P therefore have slope 5

So, just use the point-slope form of a line to write the needed equations.

Now, what is the slope of a line 45° clockwise from P? If P makes an angle Ø with the x-axis, then we want θ such that

tanθ = tan(Ø-45) = (tanØ-1)/(1+tanØ)
= (5-1)/(1+5) = 2/3

Now plug that into your point-slope form.