Find the area of the triangle with vertices \((-8,4),\ (-6,6),\ (-3,9)\).
0
Step 1: Let the vertices be \(A(-8,4)\), \(B(-6,6)\), \(C(-3,9)\).
Step 2: Use the coordinate geometry formula for the area of a triangle:
\(\text{Area} = \dfrac{1}{2}\left|x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)\right|\)
Step 3: Substitute \((x_1,y_1)=(-8,4)\), \((x_2,y_2)=(-6,6)\), \((x_3,y_3)=(-3,9)\):
\(\text{Area} = \dfrac{1}{2}\left|(-8)(6-9) + (-6)(9-4) + (-3)(4-6)\right|\)
Step 4: Simplify each term:
Step 5: Add them:
\(24 + (-30) + 6 = 0\)
Step 6: Compute the area:
\(\text{Area} = \dfrac{1}{2}|0| = 0\)
Final Answer: The area of the triangle is 0 square units (so the three points are collinear).