A rectangle is inscribed in a circle with a diameter lying along the line $$3y = x + 7$$. If the two adjacent vertices of the rectangle are $$(-8, 5)$$ and $$(6, 5)$$, then the area of the rectangle (in sq. units) is:

We begin with the two given adjacent vertices of the rectangle, $$A(-8,5)$$ and $$B(6,5)$$. Since the ordinates are equal, the segment $$AB$$ is horizontal.

Distance formula: for points $$(x_1,y_1)$$ and $$(x_2,y_2)$$ the distance is $$\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}.$$ Substituting $$A(-8,5)$$ and $$B(6,5)$$ we get $$AB=\sqrt{(6-(-8))^2+(5-5)^2}=\sqrt{14^2+0}=14.$$ So the base of the rectangle is $$14$$ units long.

Let the unknown height of the rectangle be $$h>0$$, but we do not yet decide whether the other two vertices lie above or below the segment $$AB$$. Because the sides of a rectangle are perpendicular, the two remaining vertices will share the same abscissae as $$A$$ and $$B$$ and their ordinates will differ from 5 by $$h$$.

If the rectangle rises above $$AB$$ we have $$C(6,5+h),\;D(-8,5+h).$$ If it drops below $$AB$$ we have $$C(6,5-h),\;D(-8,5-h).$$

The centre $$O$$ of a rectangle circumscribed by a circle is the intersection point of its diagonals, i.e. the midpoint of any diagonal. Using the midpoint formula $$\text{Midpoint}=\Bigl(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\Bigr),$$ we take diagonal $$AC$$ in each case.

Case 1 (vertices above): $$O=\Bigl(\dfrac{-8+6}{2},\dfrac{5+(5+h)}{2}\Bigr)=\Bigl(-1,\dfrac{10+h}{2}\Bigr).$$

Case 2 (vertices below): $$O=\Bigl(\dfrac{-8+6}{2},\dfrac{5+(5-h)}{2}\Bigr)=\Bigl(-1,\dfrac{10-h}{2}\Bigr).$$

We are told that a diameter of the circle lies along the straight line $$3y=x+7$$. A diameter’s every point, and therefore the centre of the circle itself, must satisfy the line’s equation.

So we substitute the centre’s coordinates into $$3y=x+7$$ and solve for $$h$$.

For Case 1: $$3\Bigl(\dfrac{10+h}{2}\Bigr)=-1+7=6.$$ Hence $$\dfrac{30+3h}{2}=6\;\Longrightarrow\;30+3h=12\;\Longrightarrow\;3h=-18\;\Longrightarrow\;h=-6.$$ But $$h$$, being a length, cannot be negative, so this case is impossible.

For Case 2: $$3\Bigl(\dfrac{10-h}{2}\Bigr)=6.$$ Thus $$\dfrac{30-3h}{2}=6\;\Longrightarrow\;30-3h=12\;\Longrightarrow\;3h=18\;\Longrightarrow\;h=6.$$ Here $$h$$ is positive, so this configuration is valid: the rectangle extends 6 units below the segment $$AB$$.

Finally, the area of the rectangle equals base × height: $$\text{Area}=AB\cdot h=14\times6=84\ \text{square units}.$$

Hence, the correct answer is Option D.