The area of a triangle is 5 square units. Two of its vertices are } (2,1) and (3,-2). The third vertex, which lies on the line $$y = x + 3$$​, is given by:

Area of triangle} = 5 (square units)
​$$A(2, 1),\quad B(3, -2),\quad C(x, y)\text{ lies on } y = x + 3$$​​
Formula used:
Area $$= \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|$$​​
Substitute:
​$$\frac{1}{2} \left| 2(-2 - y) + 3(y - 1) + x(1 - (-2)) \right| = 5 \\\frac{1}{2} \left| -4 - 2y + 3y - 3 + 3x \right| = 5 \\\frac{1}{2} \left| 3x + y - 7 \right| = 5 \\\left| 3x + y - 7 \right| = 10$$​​
But $$y = x + 3 \Rightarrow \\$$​​
​$$\left| 3x + (x + 3) - 7 \right| = 10 \\\left| 4x - 4 \right| = 10$$​​
Case 1: $$4x - 4 = 10 \Rightarrow x = \frac{14}{4} = \frac{7}{2},\quad y = \frac{13}{2} \\$$​​
Case 2: $$4x - 4 = -10 \Rightarrow x = \frac{-6}{4} = \frac{-3}{2},\quad y = \frac{3}{2}$$​​
Correct options: $$\left(\frac{7}{2}, \frac{13}{2}\right),\quad \left(\frac{-3}{2}, \frac{3}{2}\right)$$​​
Correct answer is (b)