Question

# A variable straight line drawn through the point of intersection of lines $$\displaystyle \frac{x}{a} + \frac{y}{b} = 1$$ and $$\displaystyle \frac{x}{b} + \frac{y}{a} = 1$$, 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)
B
2xy(ab)=ab(x+y)
C
2xy(ab)=ab(xy)
D
2xy(a+b)=ab(x+y)

Solution

## The correct option is D $$2xy (a + b) = ab(x + y)$$we have,$$\dfrac { x }{ a } +\dfrac { y }{ b } =1\quad and\quad \dfrac { x }{ b } +\dfrac { y }{ a } =1$$$$\Rightarrow bx+ay-ab=0\longrightarrow \left( 1 \right)$$$$and, \, ax+by-ab=0\longrightarrow \left( 2 \right) \\$$Equation of line through their point of intersection will be given by:$$\left( bx+ay-ab \right) +k\left( ax+by-ab \right) =0$$where $$k$$ is constant$$\Rightarrow x\left( b+ak \right) +y\left( a+bk \right) -ab\left( k-1 \right) =0$$Line meet $$x$$-axis at $$A$$, hence for $$A$$, $$y=0$$$$\Rightarrow x\left( b+ak \right) -ab\left( k-1 \right) =0$$$$\Rightarrow x=\dfrac { ab\left( k-1 \right) }{ \left( b+ak \right) }$$so, coordinate of $$A$$ is $$\left( \dfrac { ab\left( k-1 \right) }{ \left( b+ak \right) } ,0 \right)$$Line meets $$y$$-axis at $$B$$, hence for $$B$$, $$x=0$$$$\Rightarrow y\left( a+bk \right) -ab\left( k-1 \right) =0$$$$\Rightarrow y=\dfrac { ab\left( k-1 \right) }{ \left( a+bk \right) }$$so, coordinate of $$B$$ is $$\left( 0,\dfrac { ab\left( k-1 \right) }{ \left( a+bk \right) } \right)$$Let $$(h,m)$$ be midpoint of $$AB$$, hence$$h=\dfrac { \dfrac { ab\left( k-1 \right) }{ \left( b+ak \right) } +0 }{ 2 }$$$$\Rightarrow k=\dfrac { b }{ a } \left( \dfrac { 2h+a }{ b-2h } \right) \longrightarrow (3)\\ and,\quad m=\dfrac { 0+\dfrac { ab\left( k-1 \right) }{ a+bk } }{ 2 }$$$$\Rightarrow k=\dfrac { a }{ b } \left( \dfrac { 2m+b }{ a-2m } \right) \longrightarrow \left( 4 \right)$$From (3) and (4) we get$$\Rightarrow\dfrac { b }{ a } \left( \dfrac { 2h+a }{ b-2h } \right) =\dfrac { a }{ b } \left( \dfrac { 2m+b }{ a-2m } \right)$$To get equation of locus we take $$h\rightarrow x\quad and\quad m\rightarrow y$$$$\Rightarrow \dfrac { b }{ a } \left( \dfrac { 2x+a }{ b-2x } \right) =\dfrac { a }{ b } \left( \dfrac { 2y+b }{ a-2y } \right)$$$$\Rightarrow 2xy\left( a+b \right) =ab\left( x+y \right)$$Maths

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