Yes the slope of the original line is m = -2/3
Therefore the slope of the line you want is m' = -1/m =3/2
sour line is of form
y = (3/2) x + b
put in your point
4 = (3/2)(-3) + b
b= 8/2 + 9/2 = 17/2
so
y = (3/2) x + 17/2
or
2 y - 3x = 17
Determine an equation for the straight line passing through the point (-3,4) and perpendicular to the line 2x + 3y - 6 = 0.
I converted it to y=2 - (2/3)x which works fine.
This means that 0.5 is probably the right angle, to get the point at -3,4 however is the tougher part. I know it is 0.5x + (2/3)x + 7.5 .. But I got that using a graph calculator with a couple trial and errors. I don't know how to get 'b' here.
3 answers
Very simple method:
Since the new line must be perpendicular to 2x + 3y - 6 = 0,
it must take the form 3x - 2y = C
plug in (-3,4) ----> 3(-3) - 2(4) = C
C = -17
3x - 2y = -17 , all done
Since the new line must be perpendicular to 2x + 3y - 6 = 0,
it must take the form 3x - 2y = C
plug in (-3,4) ----> 3(-3) - 2(4) = C
C = -17
3x - 2y = -17 , all done
Oh that is real easy! Awesome!
3 answers