Question

# A variable line, drawn through the point of intersection of the straight lines $$\dfrac {x}{3}+\dfrac {y}{2}=1$$ & $$\dfrac {x}{2}+\dfrac {y}{3}=1$$, meets the coordinate axes in A & B. The locus of the mid point of AB is the curve

A
6(x+y)=10xy
B
(x+y)=10xy
C
xy=6(x+y)
D
2(x+y)=6xy

Solution

## The correct option is C $$6(x+y)=10xy$$The equation of line through the point os intersection of $$\displaystyle \frac { x }{ 3 } +\frac { y }{ 2 } -1=0$$ and $$\displaystyle \frac { x }{ 2 } +\frac { y }{ 3 } -1=0$$ is of the form$$\displaystyle \left( \frac { x }{ 3 } +\frac { y }{ 2 } -1 \right) +\lambda \left( \frac { x }{ 2 } +\frac { y }{ 3 } -1 \right) =0$$Putting $$y=0$$ we get point $$A$$ as $$\displaystyle \left\{ \frac { 6\left( 1+\lambda \right) }{ 2+3\lambda } ,0 \right\}$$Putting $$x=0$$ we get point $$B$$ as $$\displaystyle \left\{ 0,\frac { 6\left( 1+\lambda \right) }{ 3+2\lambda } \right\}$$If $$\left( h,k \right)$$ be the mid-point of $$AB$$, then $$\displaystyle 2h=\frac { 6\left( 1+\lambda \right) }{ 2+3\lambda }$$ and $$\displaystyle 2k=\frac { 6\left( 1+\lambda \right) }{ 3+2\lambda }$$In order to find the locus of $$\left( h,k \right)$$ we have to eliminate the variable $$\lambda$$Now $$\displaystyle \frac { 1 }{ 2h } +\frac { 1 }{ 2k } =\frac { \left( 2+3\lambda \right) +\left( 3+2\lambda \right) }{ 6\left( 1+\lambda \right) } =\frac { 5\left( 1+\lambda \right) }{ 6\left( 1+\lambda \right) }$$$$\displaystyle \Rightarrow \frac { h+k }{ 2hk } =\frac { 5 }{ 6 }$$$$\Rightarrow 6\left( h+k \right) =10hk$$Hence, the locus of mid-point $$\left( h,k \right)$$ is $$6\left( x+y \right) =10xy$$.Maths

Suggest Corrections

0

Similar questions
View More

People also searched for
View More