write the equation of the line a) parallel to b)perpendicular to 4x+3y-10=0 and is 3 units from the point (2,-1)

1 answer

given: 4x+3y - 10 = 0

a) easiest way:
since the new line is parallel it must look like
4x+3y + c = 0
using your distance formula from a point to a line
|4(2) + 3(-1) + c|/√(4^2 + 3^2) = 3
8 - 3 + c = 15
c = 10

4x + 3y + 10 = 0
or

-8 + 3 - c = 15
-c = 20
c = -20

4x + 3y - 20 = 0

We have two equations, since one is 3 units above the given point, and the other is 3 units below the given point (2,-1)

b)for perpendicular, the equation must look like
3x - 4y + k = 0
repeat as above