CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


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

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
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image