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Home Q & A Board Science-Related Homework Help Mathematics
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RJ5876 RJ5876
wrote...
Posts: 62
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11 years ago
How could I use differentiation to calculate the minimum distance the line 4x - 5y + 8 = 0 is from the origin.
I'm not sure how I could incorporate the distance formula into this either. I've tried but .. no avail.

Any assistance to this question is greatly appreciated!
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wrote...
#1
11 years ago
i donno abt the differentiation,but i can tell u a formula for this..
if ax+by+c=0 is the line and (c,d) is the point
then shortest distance is (ac+bd+c)/(a^2+b^2)^0.5
so the answer is 1.249
wrote...
#2
11 years ago
rearrange to y = (4x+8)/5
So distance(x) = sqrt{x^2 + [(4x+8)/5]^2} is to be minimised using calculus...
x = -32/41, y = 40/42 seems to be min
wrote...
#3
11 years ago
Minimum distance means shortest distance. That is perpendicular distance, It is
|4 x 0 - 5 x 0 + 8|/?(4^2 + 5^2)  = 8/?41

you dont need any calculus to do this
wrote...
#4
11 years ago
5y = - 4x - 8
y = (- 4/5) x - 8/5
m1 = - 4/5
Line from origin has m2 = 5/4
Equation is y = (5/4)x

Lines intersect at (5/4)x = (- 4/5)x - 8/5
25x = - 16x - 32
41x = - 32
x = - 32/41

y = (5/4) (-32/41)
y = - 160 / 164
y = - 40/41
d² = (32/41)² + (40/41)²
d² = (1024 + 1600) / 1681
d² = 2624/1681
d = 1.25
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