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

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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Ola, I'm not understanding this question, any assistance?