Calculus Question

Question

Find the distance between the point P (3, 1) and the line L: −2x + 3y + 6 = 0.

Answer 1

3y = 2x-6

y = (2/3)x - 2

the perpendicular line has slope -3/2

B = y - mx = 1 - (-3/2)(3)

= 1 - (-9/2)

= 11/2

y = (-3/2)x + 11/2 is the perpendicular line

they meet at (2/3)x - 2 = (-3/2)x + 11/2

multiplies by 6:

4x - 12 = -9x + 33

13x = 45

x = 45/13

(2/3)(45/13) - 2 = 30/13 - 2 = 4/13

(-3/2)(45/13) + 11/2 = -135/26 + 11/2 = -135/26 + 143/26 = 8/26 = 4/13

the intersection point is (45/13, 4/13)

the distance between (45/13) and ( 3,1) is

sqrt ( ( 3 - 45/13)^2 + ( 1 - 4/13)^2 ) =

sqrt ( ( -6/13)^2 + (9/13)^2 ) =

sqrt( 36/169 + 81/169) =

sqrt( 117/169) = sqrt ( 3 * 3 * 13/ 169) = 3*sqrt(13)/13