Answer: 10
Step-by-step explanation:
It appears that [tex]\overline{AC} \perp \overline{BC}[/tex], so to check if this is the case, we can use the slope formula.
[tex]m_{\overline{AC}}=\frac{3-1}{6-2}=\frac{1}{2}\\m_{\overline{BC}}=\frac{3-7}{6-4}=-2[/tex]
Since these slopes are negative reciprocals, we can see that this is the case. So, we can take [tex]\overline{AC}[/tex] as the base of the triangle and [tex]\overline{BC}[/tex] as the height of the triangle, and then use the formula [tex]A=\frac{1}{2}bh[/tex].
Using the distance formula,
[tex]AC=\sqrt{(6-2)^{2}+(3-1)^{2}}=\sqrt{16+4}=2\sqrt{5}\\BC=\sqrt{(4-6)^{2}+(7-3)^{2}}=\sqrt{4+16}=2\sqrt{5}[/tex]
So, the area is [tex]\frac{1}{2}(2\sqrt{5})(2\sqrt{5})=\boxed{10}[/tex]
