# A straight line L with negative slope passes through (8, 2) and cuts the positive axes at P and Q. As L varies, the absolute minimum value of OP + OQ is (O is origin) (A) 10 (B) 18 (C) 16 (D) 12

1
by lisakar98

2015-04-01T11:22:31+05:30

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Let slope = - m ,  where m > 0.
let point A be (8, 2).

equation of L :  y = - m x + c.    It passes through A.
2 = - m * 8 + c  =>  c = 2 + 8 m
OP = x intercept , ie., value of x when y = 0.
0 = - m * OP + c  = - m * OP + 2 + 8 m
=> OP = 2 / m + 8
OQ = y intercept , ie., value of y when x = 0
OQ = c = 2 + 8 m

OP + OQ = 8 m + 10 + 2 / m
derivative of (OP + OQ) wrt m :  8 - 2 / m²
derivative = 0 when:  m = +1/2  or  -1/2.  we take only the positive value.

minimum value of OP + OQ = 4 + 10 + 4 = 18
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simpler method:

The equation of L in the intercept form:  x/OP + y/OQ = 1
We are given that OP and OQ are positive.  As L passes through A (8, 2):
8/OP + 2/OQ = 1
8 OQ + 2 OP = OP * OQ
OP = 8 OQ / (OQ - 2)

The sum of intercepts :  OP + OQ = 8 OQ / (OQ - 2) + OQ

Derivative of OP + OQ wrt  OQ:  [ (OQ -2) * 8 - 8 OQ * 1 ] /(OQ - 2)²  + 1
= 1 - 16 /(OQ - 2)²
Derivative is 0  when  (OQ - 2)² = 16
OQ - 2 = +4 or -4
OQ = 6  or  -2.  we take only the positive value.
Minimum value of  OP + OQ =  8 * 6/(6-2) + 6
= 12 + 6 = 18