Line L is given by the equation 3y+2x =9 and point P has coordinates (6,-5).

Question

a. Explain why point P is not on the Line L
b. Find the gradient of line L
c. 1.) Write down the gradient of a line perpendicular to line L
2.) Find the equation of the line perpendicular to L and passing through point P.

Answer 1

3(-5) + 2(6) = what??
-15 + 12 = -3 which is NOT 9
now
y = m x + b
3y+2x =9 ---> y = (-2/3) x + 3
slope = m = -2/3

perpendicular slope = m' = -1/m = 3/2
y = m' x + b
y = (3/2) x +b
at (6,-5)
-5 = (3/2)6 + b
b = -5 - 9 = -14
y = (3/2) x -14
2 y = 3 x - 28

Answer 2

For 2a explanation:
Since P(6,-5) does not satisfy the equation, L does not pass through point P.

On the other hand, 2b ii answer would be:
2x+3y+3=0

Answer 3

For 2b ii, the answer would be different because the equation is asking for the line perpendicular to L passing through point P. The perpendicular slope is found to be 3/2, so the equation using point-slope form becomes:

y + 5 = (3/2)(x - 6)

Multiplying both sides by 2:

2y + 10 = 3x - 18

Rearranging:

3x - 2y + 28 = 0

Which is equivalent to:

2x + 3y + 3 = 0

So both answers are correct, just written in different forms.