Question

Find the equation of the line that is perpendicular to the line 2x+3y-8=0 at the point (1,2)

Answer 1

first you need to put the equation in y = mx + b form

Answer 2

y=-2x+8
-----
3

Answer 3

2x + 3y - 8 = 0
3y = -2x + 8
y = (-2x + 8) / 3
go a little further...
y = -2/3x + 8/3

Answer 4

understand so far ?

Answer 5

yes

Answer 6

the slope for this equation is - 2/3 but we are looking for the perpendicular slope. That means we are looking for the negative reciprocal of -2/3. All that means is you flip the slope over and change the sign. So the slope we need is 3/2. Do you understand, or have I lost you ?

Answer 7

no I understand that part but how do I find the rest of the line again by plugging the point I have into the equation?

Answer 8

use the point slope formula because we have the slope and the points...
y - y1 = m(x - x1)
using slope(m) = 3/2
and points (1,2) -- x1 = 1 and y1 = 2

Answer 9

now lets sub them in....
y - 2 = 3/2(x - 1)
can you solve it from here ?

Answer 10

yes

Answer 11

so what did you get ?

Answer 12

would you like me to finish it and we can compare answers ?

Answer 13

sure

Answer 14

y - 2 = 3/2(x - 1)
y - 2 = 3/2x - 3/2
y = 3/2x - 3/2 + 2
y = 3/2x - 3/2 + 4/2
y = 3/2x + 1/2
is this what you got ?

Answer 15

yes that is what I got thanks

Answer 16

anytime :)