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Analytical Geometry

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16 Nov 2009 18:57:42 IST

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A straight line L with negative slope passes through point(8,2) & cuts positive coordinate axes at p

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A straight line L with negative slope passes through point(8,2) & cuts positive coordinate axes at points P & Q.Find the absolute minimum value OP + OQ ,as L varies,, where O=origin

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Sagar Saxena

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16 Nov 2009 21:48:24 IST

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hello dear

let the line be::

x/a + y/ b = 1

8/a + 2/b = 1

C = a+ b

C = a + 2a/ (a-8)

now solve the above equation by applying the concept of maxima and minima

by differentiating it we get::

a= 4 and a= 8

check it at second differential for it maximum and minimum value

Yagyadutt Mishra

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16 Nov 2009 22:00:44 IST

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Let the slope of line be m

Then equation can be written as y= mx + c

(8,2) satisfy it..

so y = mx + c

2 = 8m + c

so c = 2-8m

HEnce line in one variable

y = mx + 2-8m

Now OP is equal to x -cordinate when y=0

SO OP = [8m - 2]/m

OQ = 2-8m

OP+OQ = 2-8m + [8m - 2]/m

OP + OQ = 2- 8m + 8 -2/m

L = 2-8m + 8 -2/m

dL/dm = 0 -8 + 0 2/m^2

So dL/dm = 0 for maximum and minimum condition

Hence 8 = 2/m^2

so m = +1/2 or -1/2.

We have to choose -1/2 ,,,,,as given in question..slope is negative

So OP+OQ = 2 +16 + 8 + 4 = 30....ans

Chitransh Gaurav

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17 Jun 2012 14:47:08 IST

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Use properTy AM>GM, and get the minimum value as 18

Chitransh Gaurav

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17 Jun 2012 14:48:30 IST

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Use properTy AM>GM, and get the minimum value as 18

NugoRama

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17 Jun 2012 22:42:07 IST

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Well, Sum of the intercepts will be minimum when both the intercepts are equal (As Chitransh said, AM>GM), so for a line to have equal intercepts as well as a negative slope, its equation is of the form x+y = c, which satisfied (8,2) here, giving c = 10, also, intercepts as 10 and 10 each. SO the minimum sum becomed 20. Surprised? So am I. P.S. Check above 3 posts too.

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