Consider $\triangle ABC$. BC lies on a line passing through $(g,f)$.
The pair of straightlines $(x+y)(x-9y)=0$ are the perpendicular bisector of sides AB and AC of $\triangle ABC$.
Find the locus of the point A.
NOTE- We only know that BC lies on the line through (g,f),but not the exact co-ordinates of B and C.
MY ATTEMPT
I tired using co-ordinate geometry and lastly concluded that I need to reflect the line passing through (g,f) on the straght lines $(x+y)=0 $ and $(x-9y)=0$ and the intersection of the reflected lines would provide me A. But the calculations would then becomes terribly frightening and the solution looks really inelegant. Is there any other way to solve this?
Assume that the lines are the perpendicular bisectors of AB and AC (your question was unclear on this point).
Let A be the point $(a,b)$. Reflect it in the two lines to get the points $(-b,-a)$ and $(\frac{40}{41}a+\frac{9}{41}b,\frac{9}{41}a-\frac{40}{41}b)$ as B and C. They must be collinear with $(g,f)$. So using the usual determinant condition we get after a little reduction $$(a+\frac{4f-5g}{8})^2+(b+\frac{4g+5f}{8})^2=\frac{41(f^2+g^2)}{64}$$ which is a circle.