The vertices of $$ riangle A B C $$ are $$ A(-4,-2), B(8,-2) $$ and C(2,6).

Question

The vertices of $$ 	riangle A B C $$ are $$ A(-4,-2), B(8,-2) $$ and C(2,6).

UpStudy AI · Accepted Answer

We are given the vertices of $$ \triangle ABC $$ as  
$$
A(-4,-2), \quad B(8,-2), \quad C(2,6).
$$

**Step 1. Compute the side lengths**

Using the distance formula  
$$
d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2},
$$
we find:

- For side $$ AB $$:  
  $$
  d_{AB} = \sqrt{(8-(-4))^2+((-2)-(-2))^2} = \sqrt{12^2+0^2} = 12.
  $$

- For side $$ BC $$:  
  $$
  d_{BC} = \sqrt{(2-8)^2+(6-(-2))^2} = \sqrt{(-6)^2+8^2} = \sqrt{36+64} = 10.
  $$

- For side $$ AC $$:  
  $$
  d_{AC} = \sqrt{(2-(-4))^2+(6-(-2))^2} = \sqrt{6^2+8^2} = \sqrt{36+64} = 10.
  $$

Thus, $$ \triangle ABC $$ is an isosceles triangle with $$ AC = BC = 10 $$.

**Step 2. Compute the area using the determinant formula**

The area of a triangle with vertices $$(x_A, y_A)$$, $$(x_B, y_B)$$, and $$(x_C, y_C)$$ is given by  
$$
\text{Area} = \frac{1}{2}\left| x_A(y_B-y_C) + x_B(y_C-y_A) + x_C(y_A-y_B) \right|.
$$

Substitute the given coordinates:
$$
\begin{aligned}
\text{Area} &= \frac{1}{2}\Big| (-4)\big[(-2)-6\big] + 8\big[6-(-2)\big] + 2\big[(-2)-(-2)\big] \Big| \
&=\frac{1}{2}\Big| (-4)(-8) + 8(8) + 2(0) \Big| \
&=\frac{1}{2}\Big| 32 + 64 \Big| \
&=\frac{1}{2}(96) \
&=48.
\end{aligned}
$$

**Final Answer**

The area of $$ \triangle ABC $$ is $$ 48 $$ square units.
The area of triangle ABC is 48 square units.

The vertices of are and
C(2,6).

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The vertices of are and C(2,6).