|
Question 169771: Write the equation of the line passing through (3,-2), but perpendicular to y=3.
I tried to use the point-slope intercept form to find the slope, but I do not have two points given. So I used the first point (3,-2) and (0,3) since y=3. Then I tried to find my slope.
y2-y1/x2-x1= 3-(-2)/0-3= -5/3
now I have a slope of -5/3
Next I tried to use the point-slope equation
y-y1=m(x-x1)
y-(-2)=-5/3(x-3)
y+2=-5/3(x-3)
3y+6=-3x+9
3y=-3x+3
Y=-x+1
Now that I have my equation is this equation perpendicular to y=3?
Do I have the right concept?
Found 2 solutions by Alan3354, solver91311:
Answer by Alan3354(69443)   (Show Source):
Answer by solver91311(24713)  (Show Source):
You can put this solution on YOUR website! You made this way too difficult on yourself.
 is a horizontal line, therefore the desired perpendicular must be a vertical line. Vertical lines have the characteristic that the x-coordinate for any ordered pair representing a point on the line is a constant value, and the y-coordinate can be any real number. Hence, if the desired line passes through (3,-2), then the x-coordinate of ALL points on the line must be 3 and therefore the equation of the desired line is 
|
|
|
| |
