|
Question 197509: Please help with the following question Write the equation for a line that is perpendicular to the line, y=3x+10/3 and goes thru the point [-15,6]
Answer by nerdybill(7384)   (Show Source):
You can put this solution on YOUR website! Please help with the following question Write the equation for a line that is perpendicular to the line, y=3x+10/3 and goes thru the point [-15,6]
.
The slope of (from inspection)
y=3x+10/3
is 3
.
If a line is to be perpendicular the slope has to be the "negative reciprocal":
Let m = our new slope
then
3m = -1
m = -1/3 (our new slope)
.
Using our new slope and [-15,6]
plug it into the "point-slope" form:
y - y1 = m(x - x1)
y - 6 = (-1/3)(x - (-15))
y - 6 = (-1/3)(x + 15)
y - 6 = (-1/3)x - 5
y = (-1/3)x + 1 (this is what they're looking for)
|
|
|
| |
