Question

Let $ABCD$ 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 $ABCD$. $L$ is the line through $A$.
A line $M$ through $A$ is drawn parallel to $BD$. 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 $AC$ 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
$y+1=-(x+1)$
$y+1=-x-1$
$\Rightarrow x+y+2=0$

$A{T}_1=\frac{OA}{2}=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}$
$A{T}_2=A{T}_2=2A{T}_1=\sqrt{2}$
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.