How do you find a standard form equation for the line perpendicular to 3x-2y+5=0 and with the same y-intercept as 3x-y+18=0?

1 Answer
Jul 23, 2016

Answer:

#2x+3y-54=0#

Explanation:

We can find slope and #y#-intercept of a line by converting its equation to point slope form. In a point-slope form of equation #y=mx+c#, #m# (the coefficient of #x# variable) is the slope and #c# is the #y#-intercept of the line.

Now, #3x-y+18=0# can be written as #y=3x+18# and hence its intercept on #y#-axis is #18#.

Further #3x-2y+5=0# can be written as #2y=3x+5# or

#y=3/2x+5/2# and hence its slope is #3/2#.

Now as product of slopes of two perpendicular lines is #-1#, slope of line perpendicular to #3x-2y+5=0# (whose slope is #3/2#) will be #(-1)/(3/2)=-1xx2/3=-2/3#.

We now need to find the equation of a line whose slope is #-2/3# and intercept on #y#-axis is #18# and using slope intercept form of equation #y=mx+c#, this should be

#y=-2/3x+18# or #3y=-2x+54# or #2x+3y-54=0#

graph{(3x-y+18)(3x-2y+5)(2x+3y-54)=0 [-32.5, 47.5, -19.36, 20.64]}