Question

# A variable straight line drawn through the point of intersection of lines xa+yb=1 and xb+ya=1, meets the coordinate axes at A and B. Then the locus of the mid-point of AB is

A
xy(a+b)=ab(x+y)
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B
2xy(ab)=ab(x+y)
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C
2xy(ab)=ab(xy)
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D
2xy(a+b)=ab(x+y)
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Solution

## The correct option is D 2xy(a+b)=ab(x+y)we have,xa+yb=1andxb+ya=1⇒bx+ay−ab=0⟶(1)and,ax+by−ab=0⟶(2)Equation of line through their point of intersection will be given by:(bx+ay−ab)+k(ax+by−ab)=0where k is constant⇒x(b+ak)+y(a+bk)−ab(k−1)=0Line meet x-axis at A, hence for A, y=0⇒x(b+ak)−ab(k−1)=0⇒x=ab(k−1)(b+ak)so, coordinate of A is (ab(k−1)(b+ak),0)Line meets y-axis at B, hence for B, x=0⇒y(a+bk)−ab(k−1)=0⇒y=ab(k−1)(a+bk)so, coordinate of B is (0,ab(k−1)(a+bk))Let (h,m) be midpoint of AB, henceh=ab(k−1)(b+ak)+02⇒k=ba(2h+ab−2h)⟶(3)and,m=0+ab(k−1)a+bk2⇒k=ab(2m+ba−2m)⟶(4)From (3) and (4) we get⇒ba(2h+ab−2h)=ab(2m+ba−2m)To get equation of locus we take h→xandm→y⇒ba(2x+ab−2x)=ab(2y+ba−2y)⇒2xy(a+b)=ab(x+y)

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