Question

The area of a triangle is 5 square units. Two of its vertices are (2, 1) and (3, -2) and the third vertex lies on y = x + 3, the third vertex is

• A. (3, 4)
• B. (5/2, 13/2)
• C. (7/2, 13/2)
• D. (1,9)

Answer 1

The correct option is C

(7/2, 13/2)

Area of the triangle = $\frac{1}{2}$[$x_1$($y_2$ - $y_3$) + $x_2$($y_3$ - $y_1$)+ $x_3$($y_1$ - $y_2$)] = 5

5 = $\frac{1}{2}$[2(-2 - y) + 3(y - 1) + 3x]

5 = $\frac{1}{2}$[-4 - 2y + 3y - 3 + 3x]

3x + y = 17, which is a linear equation in two variable and also a straight line. We are given in question that the third vertex lies on y = x + 3. It means that the point of intersection of both the lines is the vertex. SO, solving y = x + 3 and 3x + y = 17, we get x = $\frac{7}{2}$ and y = $\frac{13}{2}$.

Therefore third vertex is ($\frac{7}{2}$, $\frac{13}{2}$​)