Two vertices of a triangle are $A(2, 1)$ and $B(3, -2)$. The third vertex is $C(x, y)$,where $y = x + 3$. If the area of the triangle is $5$,find the coordinates of the third vertex.
(A) The area of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ is given by the formula: $\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.
Given vertices are $A(2, 1)$,$B(3, -2)$,and $C(x, x+3)$.
Substituting the values into the formula: $5 = \frac{1}{2} |2(-2 - (x+3)) + 3((x+3) - 1) + x(1 - (-2))|$.
$10 = |2(-x - 5) + 3(x + 2) + x(3)|$.
$10 = |-2x - 10 + 3x + 6 + 3x|$.
$10 = |4x - 4|$.
This gives two cases:
Case $1$: $4x - 4 = 10 \Rightarrow 4x = 14 \Rightarrow x = \frac{7}{2}$. Then $y = \frac{7}{2} + 3 = \frac{13}{2}$.
Case $2$: $4x - 4 = -10 \Rightarrow 4x = -6 \Rightarrow x = -\frac{3}{2}$. Then $y = -\frac{3}{2} + 3 = \frac{3}{2}$.
Thus,the coordinates of the third vertex are $(\frac{7}{2}, \frac{13}{2})$ and $(-\frac{3}{2}, \frac{3}{2})$.
Given vertices are $A(2, 1)$,$B(3, -2)$,and $C(x, x+3)$.
Substituting the values into the formula: $5 = \frac{1}{2} |2(-2 - (x+3)) + 3((x+3) - 1) + x(1 - (-2))|$.
$10 = |2(-x - 5) + 3(x + 2) + x(3)|$.
$10 = |-2x - 10 + 3x + 6 + 3x|$.
$10 = |4x - 4|$.
This gives two cases:
Case $1$: $4x - 4 = 10 \Rightarrow 4x = 14 \Rightarrow x = \frac{7}{2}$. Then $y = \frac{7}{2} + 3 = \frac{13}{2}$.
Case $2$: $4x - 4 = -10 \Rightarrow 4x = -6 \Rightarrow x = -\frac{3}{2}$. Then $y = -\frac{3}{2} + 3 = \frac{3}{2}$.
Thus,the coordinates of the third vertex are $(\frac{7}{2}, \frac{13}{2})$ and $(-\frac{3}{2}, \frac{3}{2})$.
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