+1671 Please help with this: Find the distance between (3,4) and the line 4x+3y+7=2x-y+5.
+128407 Rearrange the equation in the form Ax + By + C = 0
2x + 4y + 2 = 0
We have the following for the distance from (x, y) to the line =
l Ax + By + C l l 2 (3) + 4 ( 4) + 2 l 24 24
________________ = ___________________ = ______ = ______ =
sqrt [ A^2 + B^2] sqrt [ 2^2 + 4^2 ] sqrt 20 2 sqt 5
12 / sqrt 5 =
(12/5) sqrt (5)
See the graph here : https://www.desmos.com/calculator/3dgerbkzta
+36915 Another (but longer) method ( because I can never remember the equation for the minumum distance between a line and a point)
4x+3y+7=2x-y+5 re-arrange to y = -1/2 x -1/2
Now use distance formula
d^2 = (x -3)^2 + (4 - (-1/2x-1/2))^2
x^2 -6x + 9 + 1/4x^2 + 4.5 x + 20.25
d^2 = 5/4 x^2 - 1.5 x + 29.25 <====== take derivative and set = 0
10/4 x -1.5 = 0
x = 0.6 <====== sub this value into the line equation to find y = - 0.8
NOW, use the distance formula between (0.6, -0.8) and (3,4)
d^2 = ( 3 - 0.6)2 + (4- - 0.8)2
d^2 = 28.8
d = sqrt (28.8) ( which = same answer that chris found....)
