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  • A variable line drawn through the point of intersection of the straight lines <img alt="\mathrm{\frac{x}{a}+\frac{y}{b}=1 \text\ { and }\ \frac{x}{b}+\frac{y}{a}=1}" src="https://entrancecorne

A variable line drawn through the point of intersection of the straight lines \mathrm{\frac{x}{a}+\frac{y}{b}=1 \text\ { and }\ \frac{x}{b}+\frac{y}{a}=1} meets the co-ordinate axes in \mathrm{A} & \mathrm{B}. Find the locus of the mid point of \mathrm{AB}. 

 

Option: 1

\mathrm{\frac{x+y}{a b}=\frac{2 a b}{x y}}


Option: 2

\mathrm{\frac{x y}{a b}=\frac{x-y}{a-b}}


Option: 3

\mathrm{x y(a+b)=2 a b(x+y)}


Option: 4

\mathrm{\frac{a b}{x y}=\frac{2 a b}{x+y}}


Answers (1)

best_answer

\mathrm{\begin{aligned} & \frac{x}{a}+\frac{y}{b}=1 \ \ \ \ \ \ .......\left ( i \right ) \\ & \frac{x}{b}+\frac{y}{a}=1 \ \ \ \ \ \ ........\left ( ii \right ) \end{aligned}}
\mathrm{P} is the point of intersection line \mathrm{\left ( i \right )} is reflection of the line \mathrm{\left ( ii \right )} about \mathrm{y=x} they must intersect at \mathrm{y=x}
\mathrm{\frac{x}{a}+\frac{y}{b}=1}                    \mathrm{\Rightarrow x=\frac{a b}{a+b}=y}

\mathrm{\therefore } Point \mathrm{P } \mathrm{\left(\frac{a b}{a+b}, \frac{a b}{a+b}\right)} a variable line passes through point \mathrm{P} cuts the coordinate axes in \mathrm{A} & \mathrm{B}. If mid point of \mathrm{AB} is \mathrm{J(h, k)} then co-ordinates of \mathrm{A} and \mathrm{B} are \mathrm{(2h, 0)} & \mathrm{ (0, 2k)}
\mathrm{\therefore } equation to line \mathrm{ AB} is \mathrm{ \frac{x}{2 h}+\frac{y}{2 k}=1}
\mathrm{ \because } \mathrm{ P } lies on this line 
\mathrm{ \begin{aligned} & \left(\frac{a b}{a+b}\right) / 2 h+\frac{(a b / a+b)}{2 k}=1 \quad \Rightarrow \frac{1}{2}\left(\frac{a b}{a+b}\right)\left(\frac{h+k}{h k}\right)=1 \quad \Rightarrow 2 a b(h+k)=h k(a+b) \\ & \Rightarrow x y(a+b)=2 a b(x+y) \end{aligned} }


 

 

Posted by

Rishi

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