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# a straight line L with negative slope passes through the point (8,2) and cuts positive coordinate axes at the points P and Q. Find the minimum value of OP + OQ as L varies where .

Vikas TU
14149 Points
5 years ago
a straight line L with negative slope having let intercepts with x  and y respectively be:
a and b.
Therefore, eqn. of the line passing through (8,2) is:
8/a + 2/b = 1
or
a = 8b/b-2 …...................(1)

minimum value of OP + OQ
is: a + b = M (let)
from  eqn. (1)
M = b + 8b/b-2
dM/db = 1 + 8{(b-2) – (b)}/(b-2)^2 = 0
b^2 + 3b – 12 = 0
(b-3) (b + 4) = 0
b =3 since it is on positive y axis.
put y in eqn. (1)
u will get,
M = 27 minimum value. hennce!
Bob
13 Points
4 years ago
from point slope form we have:
let the slope of ine L be -m.
y1 = 2, x1 = 8.
y-2 = -m(x-8)

y -2 = -mx + 8m
mx + y = 8m + 2

mx/(8m+2)/m + y/(8m+2) = 1

therefore, OP = (8m + 2)/m    OQ = 8m + 2

Let S = OP + OQ

(8m + 2)/m + 8m + 2

8 + 2/m + 2 + 8m

10 + 2/m + 8/m

using completing the square method we get:

18 + [ (23/2m1/2 – (2/m)1/2  ]2

For minimum value of S
[ (23/2m1/2 – (2/m)1/2  ]2   =  0

solving we get m= ½
and m= -1/2

So, minimum value of OP + OQ will be 18 (for m=-1/2 i.e negative slope)