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😃: area of triangle 5-12-13
🤖: To find the area of a triangle with vertices at coordinates $A(5, 1)$, $B(1, 1)$, and $C(-3, 1)$, we can use the formula for the area of a triangle given its vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$:

$$\text{Area}=\frac{1}{2}\mid x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)+x_1(y_2-y_3)\mid \backslash text\{Area\}\; =\; \backslash frac\{1\}\{2\}\; \backslash left|\; x\_1(y\_2\; -\; y\_3)\; +\; x\_2(y\_3\; -\; y\_1)\; +\; x\_3(y\_1\; -\; y\_2)\; +\; x\_1(y\_2\; -\; y\_3)\; \backslash right|$$

Substituting the coordinates of points $A(5, 1)$, $B(1, 1)$, and $C(-3, 1)$ into the formula, we get:

$$\text{Area}=\frac{1}{2}\mid 5(1-1)+1(1-1)+(-3)(1-1)+5(1-1)\mid \backslash text\{Area\}\; =\; \backslash frac\{1\}\{2\}\; \backslash left|\; 5(1\; -\; 1)\; +\; 1(1\; -\; 1)\; +\; (-3)(1\; -\; 1)\; +\; 5(1\; -\; 1)\; \backslash right|$$

Simplify inside the absolute value:

$$\text{Area}=\frac{1}{2}\mid 5\cdot 0+1\cdot (-2)+(-3)\cdot 0+5\cdot 0\mid =\frac{1}{2}\mid 0-2+0+0\mid =\frac{1}{2}\mid -6\mid =\frac{1}{2}\cdot 6=3\backslash text\{Area\}\; =\; \backslash frac\{1\}\{2\}\; \backslash left|\; 5\; \backslash cdot\; 0\; +\; 1\; \backslash cdot\; (-2)\; +\; (-3)\; \backslash cdot\; 0\; +\; 5\; \backslash cdot\; 0\; \backslash right|\; =\; \backslash frac\{1\}\{2\}\; \backslash left|\; 0\; -\; 2\; +\; 0\; +\; 0\; \backslash right|\; =\; \backslash frac\{1\}\{2\}\; \backslash left|\; -6\; \backslash right|\; =\; \backslash frac\{1\}\{2\}\; \backslash cdot\; 6\; =\; 3$$

Thus, the area of the triangle is $\boxed{6}$.]

The area of the triangle is $\boxed{6}$.]

The area of triangle $ABC$...