SOLUTION: write an equation in slope intercept form which is perpendicular to the line y= -3x+0, and goes through the midpoint (-1/2, 3/2).

Question 508812: write an equation in slope intercept form which is perpendicular to the line y= -3x+0, and goes through the midpoint (-1/2, 3/2).
Found 2 solutions by oberobic, stanbon:
Answer by oberobic(2304) (Show Source): You can put this solution on YOUR website!
y = -3x
.
The slope is -3.
.
The slope of a perpendicular line is the inverse reciprocal: 1/3.
.
So, the equation is
.
y = 1/3x + b
.
To find 'b' we can use the point we want the line to go through.
.
(-1/2,3/2) = (x, y)
.
3/2 = 1/3(-1/2) + b
.
3/2 = -1/6 + b
b = 9/6 + 1/6
b = 10/6
b = 5/3
.
y = 1/3x + 5/3
.
The following graph shows the y = -3x in red and y = 1/3x+5/3 in green
.

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
write an equation in slope intercept form which is perpendicular to the line y= -3x+0, and goes through the midpoint (-1/2, 3/2).
-------------
The slope of y = -3x is -3
---
The slope you want is 1/3
---
Form: y = mx+b
3/2 = (1/3)(-1/2) + b
---
3/2 = (-1/6) + b
---
9/6 = (-1/6) + b
b = 10/6
b = 5/3
-----
Equation:
y = (1/3)x + (5/3)
========================
Cheers,
Stan H.
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