The area of a triangle is 5 sq units. Two of its vertices are (2, 1) and (3, -2). The third vertex lies on y = x + 3. Find the third vertex.
Answer: D
Let the third vertex be (x, y). Since it lies on y = x+3, its coordinates are (x, x+3).
Using the area formula: Area = \(\frac{1}{2} |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|\)
5 = \(\frac{1}{2} |2(-2-(x+3)) + 3((x+3)-1) + x(1-(-2))|\)
10 = |2(-x-5) + 3(x+2) + x(3)|
10 = |-2x - 10 + 3x + 6 + 3x| = |4x - 4|
This gives two possibilities:
1) 4x - 4 = 10 => 4x = 14 => x = 14/4 = 7/2. Then y = 7/2 + 3 = 13/2. Vertex is (7/2, 13/2).
2) 4x - 4 = -10 => 4x = -6 => x = -6/4 = -3/2. Then y = -3/2 + 3 = 3/2. Vertex is (-3/2, 3/2).