Question

Let $A B C D$ be a square of side length $2$ units. $C_{2}$ is the circle through vertices $A, B, C, D$ and $C_{1}$ is the circle touching all the sides of square $A B C D$. $L$ is the line through $A$.
A line $M$ through $A$ is drawn parallel to $B D$. Points $S$ moves such that its distances from the line BD and the vertex $A$ are equal. If locus of $S$ cuts $M$ at $T_{2}$ and $T_{3}$ and $A C$ at $T_{1}$, then area of $\Delta T_{1} T_{2} T_{3}$ is

• A. 12 sq. unit
• B. 23 sq. unit
• C. 1 sq. unit
• D. 2 sq. unit

Answer 1

Answer: (C)

The line parallel to $BD$ that passes through point $A$ is
$y+1=-(x+1) $
$y+1=-x-1$
$\Rightarrow x+y+2=0$

$A T_{1}=\frac{O A}{2}=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}} $
$A T_{2}=A T_{2}=2 A T_{1}=\sqrt{2}$
The locus of $S$ is a parabola.
Therefore, the area of $\Delta T_{1} T_{2} T_{3}$ is
$\frac{1}{2} \times \frac{1}{\sqrt{2}} \times 2 \sqrt{2}=\frac{1}{2} \times 2=1$ sq. unit.