Okay so I'm really having a hard time finding the locus of curves or point of intersection of curves.
The problem : I am able to get the geometrical condition in most of the questions but then I have no idea what to do to get the locus.
The question :A variable straight line drawn through the point of intersection of the straight line $x+2y-4=0$ and $2x+y-4=0$ meet the coordinate axes in $A$ and $B$.Then what is the locus of midpoint of $AB$ ?
My try : I found the equation of the variable straight line as $3x+3y-8=0$,and the required mid-point in the question to be $(4/3,4/3)$.How do I proceed from here ?
Best Answer
Th equation of any line passing through the intersection of $$x+2y-4=0,2x+y-4=0$$ is
$$0=x+2y-4+m(2x+y-4)=x(1+2m)+y(2+m)-(4m+4)$$ where $m$ is an arbitrary constant
$$\implies\dfrac x{\dfrac{4m+4}{2m+1}}+\dfrac y{\dfrac{4m+4}{m+2}}=1$$
So, $A:\left(\dfrac{4m+4}{2m+1},0\right)$ and $B;\left(0,\dfrac{4m+4}{m+2}\right)$
Now if $P(h,k)$ is the midpoint,
we have $$h=\dfrac{\dfrac{4m+4}{2m+1}+0}2\iff m=?$$
$$k=\dfrac{0+\dfrac{4m+4}{m+2}}2\iff m=?$$
Equate the values of $m$ to eliminate $m,$ the variable we have introduced.
Can you take it from here?