What is the equation of the line that passes through (–2, –3) and is perpendicular to 2x – 3y = 6?

Question

shoba · Answer

first
 2x - 3y = 6
3y = 2x - 6
y = 2/3x - 2
y=mx + c

from this equation the slope ism= 2/3 and the intercept is c = -2

next find the equation of the line that parallel to y = 2/3x - 2 which pass through the point (-2,-3)

can u solve it now?

shoba · Answer

you need to use point-slope formula...

$$y - y _{1} = m(x-x _{1}) m=is  the slope and (x _{1},y _{1}) is the given point$$

shoba · Answer

m1m2 = -1 (if it is perpendicular)
 m1 = 2/3
m2 = - 3/2 (new)

x1,y1 = (-2,-3)

so 
y + 3 = -3/2 (x+2)
2y + 6 = -3x -6
2y = -3x - 12
y = -3/2 x -6