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A variable straight line is drawn throug...

A variable straight line is drawn through the point of intersection of the straight lines `x/a+y/b=1` and `x/b+y/a=1` and meets the coordinate axes at `A` and `Bdot` Show that the locus of the midpoint of `A B` is the curve `2x y(a+b)=a b(x+y)`

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The equation of the variable line through the point of intersection of the given lines is
`((x)/(a)+(y)/(b))+K((x)/(b)+(y)/(a))=0`
`or ((1)/(a)+(K)/(b ))x+((1)/(b)+(K)/(a))x+((1)/(b)+(K)/(a))y=1+K`
or `(aK+b)x+(bK+a)y=ab(1+k).`
This line cuts x-axis at `A((ab(1+K))/(aK+b),0)` and y-axis at `B(0,(ab(1+K))/(bK+a))`
Let `P(x_(1),y_(1))` be the mid point of AB.
`thereforex_(1)(ab(1+K))/(2(ak+b))and y_(1)=(ab(1+k))/(2(bk+a))`
`implies(1)/x_(1)=(2(aK+b))/(ab(1+K))and (1)/(y_(1))=(2(bK+a))/(ab(1+K))`
`(1)/(x_(1))+(1)/(y_(1))=(2(a+b)(1+k))/(ab(1+k))`
`implies(x_(1)+y_(1))/(x_(1)y_(1))=(2(a+b))/(ab)`
`impliesab(x_(1)+y_(1))=2(a+b)(x_(1)y_(1))`
Hence the locus of `P(x_(1),y_(1))is 2xy(a+b)=ab(x+y)`
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