L1 passes through (-3,-1) and is perpendicular to L2: 3x+2y+3=0. L1 equation for y=mx+b would be what?

Question

Solve for m and b, they must be in fraction form. L1 means line 1 and L2 means line 2.

Answer 1

The slope of L2 m2 = -3 / 2

when L1 and L2 are perpendicular

m1 ( slope of L1 ) = 2 / 3

then the slope of L2 = 2/3 and passing through ( -3 , - 1 )

the equation of L2 is
y + 1 = 2/3 ( x + 3 )
y + 1 = 2/3 x + 2 multiply by 3
3y + 3 = 2x + 6
2x + 6 - 3y - 3 = 0
2x - 3y + 3 = 0

Answer 2

L1 is perpendicular to L2.
Slope of L2 = -3/2
Slope of perpendicular line is negetive reciprocal ,
so , Slope of L1 = 2/3
So , Equation of L1 => As it passes through ( -3,-1 )

( y +1 ) = ( 2/3 ) (x+3)
y = (2/3)x + 2 - 1
y = (2/3)x + 1

Answer 3

To find it perpendicular, we must find the slope of the given line:

3x+2y+3=0
>2y=-3x+3
>y=(-3/2)x+(3/2)

so the slope is (-3/2) and the slope of a line perpendicular to that is (2/3); the reciprocal

given the point (-3,1); we know that the line we need must pass though 1 on the y-axis so the equation in slope-intercept form (y=mx+b) is:

y=(2/3)x+1

Answer 4

Line L2
2y = - 3x - 3
y = (- 3 / 2).x - 3 / 2
Line 1
gradient = 2 / 3
y + 1 = (2/3).(x + 3)
y = (2/3).x + 1