Asked by Mads

Write the equation of a line perpendicular to 2x-3y = -12 that passes through (6,-1)
Answered by oobleck
the slope of 2x-3y = -12 is 3/2
so your line must have slope -2/3
Now use the point-slope form of the line:

y+1 = -2/3 (x-6)
Answered by henry2,
2x-3y = -12.
m = -A/B = -2/-3 = 2/3.
b = C/B = -12/-3 = 4 = y-int.

Y = mx+b.
m = -3/2.
b = 4.
Y = (-3/2)x+4. Slope-int. form.
3x/2+y = 4
3x+2y = 8. Std. form.

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