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Question: A (a, 0), B (b, 0), C (c, 0) and D(d, 0) are four given points. If\(\frac{CA}{CB} + \frac{DA}{DB}\) ...

A (a, 0), B (b, 0), C (c, 0) and D(d, 0) are four given points. IfCACB+DADB\frac{CA}{CB} + \frac{DA}{DB} = 0, then-

A

1a+1b=1c+1d\frac{1}{a} + \frac{1}{b} = \frac{1}{c} + \frac{1}{d}

B

(a + b) (c + d) = 2 (ab + cd)

C

(a + b) ab = (c + d) cd

D

None of these

Answer

(a + b) (c + d) = 2 (ab + cd)

Explanation

Solution

We have

CACB+DADB\frac{CA}{CB} + \frac{DA}{DB}= 0 i.e. ∣c−ab−c∣+∣d−ab−d∣\left| \frac{c - a}{b - c} \right| + \left| \frac{d - a}{b - d} \right|= 0

i.e. c−ab−c±d−ab−d\frac{c - a}{b - c} \pm \frac{d - a}{b - d} = 0

Taking +ve sign, we have

(a + b) (c + d) = 2(ab + cd) Taking –ve sign, we have (a – b) (c – d) = 0

Which is not possible if the four points are distinct.