SOLUTION: Write qn equation of the line that passes through (9, 6) and is perpendicular to the line whose equation is y=-1/3x+7.
So far I have:
y=3x7/3
y-(6)=7/3 (x-9)
y+6=7/3x3
Thi
Algebra.Com
Question 98151: Write qn equation of the line that passes through (9, 6) and is perpendicular to the line whose equation is y=-1/3x+7.
So far I have:
y=3x7/3
y-(6)=7/3 (x-9)
y+6=7/3x3
This is as far as I can go. I am trying to assist my niece and at the age of 55, my brain is not what it use to be.
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
Write an equation of the line that passes through (9, 6) and is perpendicular to the line whose equation is y =-(1/3)x + 7.
:
The relationship for the slopes of perpendicular lines is: m1*m2 = -1
m1 = -1/3, find m2
(-1/3)*m2 = -1
multiply by -1
m2 = +3; it looks like you got that part OK
:
Use the point slope equation: y - y1 = m(x - x1)
:
In this equation: m = +3; x1 = 9; y1 = 6
:
y - 6 = 3(x - 9)
y - 6 = 3x - 27
y = 3x - 27 + 6
y = 3x - 21 is perpendicular to y = -(1/3)x + 7
:
Check by substituting 9 for x in the above equation
y = 3(9) - 21
Y = 27 - 21
Y = 6
:
Should look like this:
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