Q: Two fixed points A and B are taken on the co-ordinate axes such that OA = a and OB = b. Two variable points A’ and B’ are taken on the same axes such that OA’+ OB’ = OA + OB. Find the locus of the point of intersection of AB’ and A’B.

Sol: A ≡ (a,0) , B ≡ (0 , b) , A’ ≡ (a’ ,0) , B’ ≡ (0 , b’)

Equation of A’B is $\displaystyle \frac{x}{a’} + \frac{y}{b} = 1$ ….(i)

and the equation of AB’ is $\displaystyle \frac{x}{a} + \frac{y}{b’} = 1$ ….(ii)

Subtracting (i) from (ii), we get,

$\displaystyle x(\frac{1}{a} – \frac{1}{a’}) + y(\frac{1}{b’} – \frac{1}{b}) = 0 $

$\displaystyle \frac{x(a’-a)}{a a’} + \frac{y(b-b’)}{b b’} = 0 $

Using a’ – a = b – b’

$\displaystyle \frac{x}{a a’} + \frac{y}{b b’} = 0 $

$\displaystyle b’ = \frac{a(a+b)y}{ay – bx} $ …(iii)

From (ii) b’x + ay = ab’

$\displaystyle b’ = \frac{a y}{a – x} $ …(iv)

Equating (iii) and (iv) we get x + y = a + b which is the required locus

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