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find the POINT on the line 6x+7y-5=0 which is closest to the point (2,2)Asked by Taylor
Find the point on the line –3x+4y–5=0 which is closest to the point (0–5)
Answers
Answered by
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.
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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