The two vertices of a triangle are (2, -1) an | vedclass

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(C) Since the third vertex $C(x_1, y_1)$ lies on the line $x + y = 5$,we have $y_1 = 5 - x_1$.Thus,the coordinates of $C$ are $(x_1, 5 - x_1)$.Given t

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$(5, 0)$ or $(1, 4)$

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(C) Since the third vertex $C(x_1, y_1)$ lies on the line $x + y = 5$,we have $y_1 = 5 - x_1$.Thus,the coordinates of $C$ are $(x_1, 5 - x_1)$.Given the area of $\Delta ABC = 4$,we use the determinant formula for the area of a triangle:$\frac{1}{2} |x_A(y_B - y_C) + x_B(y_C - y_A) + x_C(y_A - y_B)| = 4$$\frac{1}{2} |2(2 - (5 - x_1)) + 3((5 - x_1) - (-1)) + x_1(-1 - 2)| = 4$$|2(x_1 - 3) + 3(6 - x_1) - 3x_1| = 8$$|2x_1 - 6 + 18 - 3x_1 - 3x_1| = 8$$|12 - 4x_1| = 8$This gives two cases:Case $1$: $12 - 4x_1 = 8 \implies 4x_1 = 4 \implies x_1 = 1$. Then $y_1 = 5 - 1 = 4$. So,$C$ is $(1, 4)$.Case $2$: $12 - 4x_1 = -8 \implies 4x_1 = 20 \implies x_1 = 5$. Then $y_1 = 5 - 5 = 0$. So,$C$ is $(5, 0)$.Therefore,the third vertex is $(5, 0)$ or $(1, 4)$.

The two vertices of a triangle are $(2, -1)$ and $(3, 2)$,and the third vertex lies on the line $x + y = 5$. If the area of the triangle is $4$ square units,then the third vertex is:

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