Given Information:
- $\(\triangle P Q R\)$ is a right triangle (right angle at Q ).
- X is the midpoint of PQ .
- Y is the midpoint of PR.
- T is any point on QR .
- $\(\mathrm{PQ}=6\)$
- $\(\mathrm{QR}=8\)$
1. Find the Area of the large triangle $\(\triangle P Q R\)$ :
- Area of a right triangle $\(=\frac{1}{2} \times\)$ base $\(\times\)$ height
- Area of $\(\triangle P Q R=\frac{1}{2} \times P Q \times Q R=\frac{1}{2} \times 6 \times 8=\frac{1}{2} \times 48=24\)$.
2. Properties of Midpoints (Midpoint Theorem):
- Since X is the midpoint of PQ and Y is the midpoint of $\(\mathrm{PR}, \mathrm{XY}\)$ is parallel to QR and $\(X Y=\)$ $\(\frac{1}{2} Q R\)$.
- $\(X Y=\frac{1}{2} \times 8=4\)$.

3. Area of $\(\triangle P X Y\)$ :
- Since X and Y are midpoints, $\\(triangle P X Y\)$ is similar to $\(\triangle P Q R\)$.
- The ratio of their sides is $1: 2$.
- The ratio of their areas is $\((1: 2)^2=1: 4\)$.
- Area of $\(\triangle P X Y=\frac{1}{4} \times\)$ Area of $\(\triangle P Q R=\frac{1}{4} \times 24=6\)$.
4. Area of $\(\triangle Q T R\)$ :
- This approach is not straightforward because T is "any point" on QR.
- Let's focus on the quadrilateral PXTY.
- Area of quadrilateral PXTY = Area of $\(\triangle P Q R\)$ - Area of $\(\triangle Q X T\)$ - Area of $\(\triangle T Y R\)$. This also seems complicated because T's position is arbitrary.

Let's use a different decomposition for the quadrilateral PXTY:
Area of quadrilateral PXTY $=$ Area of $\(\triangle P X Y+\)$ Area of $\(\triangle X T Y\)$.
We already found Area of $\(\triangle P X Y=6\)$.
Now we need Area of $\(\triangle X T Y\)$.
- The base of $\(\triangle X T Y\)$ can be XY . We know $\(X Y=4\)$.
- The height of $\(\triangle X T Y\)$ with respect to base XY would be the perpendicular distance from T to the line XY.
- Since XY is parallel to QR , the distance between XY and QR is constant.
- The line $X Y$ is exactly halfway between $P$ and $Q R$. The total height from $P$ to $Q R$ is $P Q=6$.
- So the distance from X to QR is $\(\frac{1}{2} P Q=\frac{1}{2} \times 6=3\)$. This is incorrect.
- The distance from P to XY is $\(\frac{1}{2} \times\)$ height of $\(\mathrm{PQR}=\frac{1}{2} \times 6=3\)$.
- The height of the trapezoid XQRY (formed by $\(\mathrm{X}, \mathrm{Q}, \mathrm{R}, \mathrm{Y}\)$ ) is PQ - (height of PXY from P to XY ) which is $6-3=3$. No, the height of the trapezoid XQRY with parallel sides XY and QR is the perpendicular distance between them.

- Let $Q$ be the origin $(0,0)$. $P$ is $(0,6), R$ is $(8,0)$.
- Since $X$ is the midpoint of $P Q, X$ is $(0,3)$.
- Since $Y$ is the midpoint of $P R, Y$ is $\(\left(\frac{0+8}{2}, \frac{6+0}{2}\right)=(4,3)\)$.
- So, XY is a horizontal line segment at $\(\mathrm{y}=3\)$.
- T is any point on QR . The line QR passes from $(0,0)$ to $(8,0)$. So T is $(t, 0)$ for $0 \leq t \leq 8$.
- The base of $\(\triangle X T Y\)$ is XY . The length $\(X Y=4\)$.
- The height of $\(\triangle X T Y\)$ with respect to base XY is the perpendicular distance from T to the line containing $X Y$.
- The line containing XY is $y=3$.
- Since T is on QR (the x -axis, $y=0$ ), the perpendicular distance from T to the line $y=3$ is simply $3-0=3$.
- Area of $\(\triangle X T Y=\frac{1}{2} \times\)$ base $\(X Y \times\)$ height (distance from T to XY) $\(=\frac{1}{2} \times 4 \times 3=\)$ 6.
5. Calculate the Area of Quadrilateral PXTY:
- Area of PXTY $=$ Area of $\(\triangle P X Y+\)$ Area of $\(\triangle X T Y\)$
- Area of PXTY $\(=6+6=12\)$.

Comparing Quantity A and Quantity B:
Quantity A: Area of quadrilateral PXTY $=12$
Quantity B: 12

Therefore, Quantity A is equal to Quantity B.
The position of T on QR does not affect the area of $\(\triangle X T Y\)$ because its height from T to base XY remains constant at 3 .

The final answer is The two quantities are equal.