SOLUTION: What is an equation of the line that passes through the point (−2,−4) and is perpendicular to the line x+5y=15. Express your answer in slope-intercept form.
y=mx+b
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-> SOLUTION: What is an equation of the line that passes through the point (−2,−4) and is perpendicular to the line x+5y=15. Express your answer in slope-intercept form.
y=mx+b
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Question 552055: What is an equation of the line that passes through the point (−2,−4) and is perpendicular to the line x+5y=15. Express your answer in slope-intercept form.
y=mx+b
x+15=5y
15x/5
y=3 Answer by mananth(16946) (Show Source):
5 y = -1 x + 15
Divide by 5
y = -1/5 x + 3
Compare this equation with y=mx+b
slope m = - 1/5
The slope of a line perpendicular to the above line will be the negative reciprocal =5
m1*m2=-1
m= 5 ,point ( -2 , -4 )
Find b by plugging the values of m & the point in
y=mx+b
-4 = -10.00 + b
b= 6
m= 5
The required equation is y = 5 x + 6
m.ananth@hotmail.ca
