SOLUTION: The equation of a line perpendicular to the given line but passing through the given point y=-2/3x+2 (9,-3). My answer is y = (3/2)x - (33/2)but I don't know how to graph this beca

Algebra ->  Linear-equations -> SOLUTION: The equation of a line perpendicular to the given line but passing through the given point y=-2/3x+2 (9,-3). My answer is y = (3/2)x - (33/2)but I don't know how to graph this beca      Log On


   



Question 855083: The equation of a line perpendicular to the given line but passing through the given point y=-2/3x+2 (9,-3). My answer is y = (3/2)x - (33/2)but I don't know how to graph this because its to big, can it be reduced? My graph only goes to -4

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
Your line does not make a nice graph.
I like to include the origin, and pieces of the x-axis and y-axis to both sides.
In this case, including the y-intercept (0,-33/2) would be difficult.
Even including point (9,-3) may be tricky.
You only need two points (or one point and the slope) to draw a line, and you could use any point(s) for your line.
in line y=%283%2F2%29x-33%2F2 , for x=9 --> y=-3 (point (9,-3) as required),
and a slope of 3%2F2 means that
as x increases by 2 , y increases by 3 .
That gives us points (11,0), (13,3), (15,6).
For line y=%28-2%2F3%29x%2B2 we could use point (0,2), and a slope of -2%2F3 would take us to (3,0), (6,-2), and (9,-5)