# calculus

posted by Taylor

Find the point on the line –3x+4y–5=0 which is closest to the point (0–5)

1. Steve

the slope of your line is 3/4
So, the slope of the perpendicular line is -4/3

The line through (0,-5) with slope -4/3 is

y+5 = -4/3 x

The two lines intersect at (-3,-1)

Or, if you want to exercise your calculus, the distance from (0,-5) to any point (x,y) is

d^2 = x^2 + (y+5)^2

Since y = (3x+5)/4,

d^2 = x^2 + ((3x+5)/4+5)^2
= 25/16 (x^2+6x+25)

2d dd/dx = 25/16 (2x+6)
dd/dx=0 when x = -3, as above.

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