SOLUTION: How do I write the equation of a line in slope-intercept form and standard form containing point (5,1) and is perpendicular to y=5x-4?

Algebra ->  Linear-equations -> SOLUTION: How do I write the equation of a line in slope-intercept form and standard form containing point (5,1) and is perpendicular to y=5x-4?      Log On


   

Question 354225: How do I write the equation of a line in slope-intercept form and standard form containing point (5,1) and is perpendicular to y=5x-4?
Answer by nerdybill(7384) About Me  (Show Source):
You can put this solution on YOUR website!
How do I write the equation of a line in slope-intercept form and standard form containing point (5,1) and is perpendicular to y=5x-4?
.
The slope of:
y=5x-4
is 5
we know this because it is in the "slope-intercept" form:
y=mx+b
.
Our new line is perpendicular to this so it must be
5m = -1
m = -1/5 (negative reciprocal)
.
With (5,1) and the slope (-1/5) plug it into "point-slope" form:
y - y1 = m(x - x1)
y - 1 = (-1/5)(x - 5)
y - 1 = (-1/5)x + 1
y = (-1/5)x + 2 (slope-intercept form)
.
standard form:
(1/5)x + y = 2