Question

Find the equation of the line that is perpendicular to the line formed by the equation 3x + 2y = -14 and passes through the point (-48, 3).

Answer 1

first change the equation to the standard form of
y = mx + b where m is the slope of the line

2y = -3x - 14

3
y = - ----x - 7
2
now it is in the standard form ans so we get slope i.e. m = -3/2

if m2 is the slope of the perpendicular line, then
1 1
m2 = - ------ = - ----- = 2/3
m -3/2

so the perpendicular line will be of the form y = m2x + b
2
=> y = ---x + b ---------------(1)
3
we r given the point on the perpendicular line as (-48,3) i.e. x = -48 and y = 3
substituting the values in (1) we get
2
=> 3 = ---- (-48) + b
3
Multiplying by 3 throughout

=> 9 = 2(-48) + 3b
=> 9 = -96 + 3b
=> 9 + 96 = 3b
=> 105/3 = b
=> 35 = b
Putting this in (1), equation of perpendicular line is
2
y = ---x + 35
3