The triangle is isosceles.
A triangle is a geometric figure having three sides. Base on the length of its sides a triangle can be classified as:
1. Equilateral: Three equal sides.
2. Isosceles: Two equal sides.
3. Scalene: No equal sides
The distance formula is:
[tex]d=\sqrt{(x_{1}-x_{2})^2+(y_{1}-y_{2})^2}[/tex]
Then:
[tex]\bullet \ For \ \overline{AB}: \\ \\ A(7, -2) \\ B(-4, 9) \\ \\ d_{AB}=\sqrt{(7-(-4))^2+(-2-9)^2} \\ \\ d_{AB}=\sqrt{(7+4)^2+(-2-9)^2} \\ \\ d_{AB}=\sqrt{(11)^2+(-11)^2} \\ \\ \boxed{d_{AB}=11\sqrt{2}}[/tex]
[tex]\bullet \ For \ \overline{AC}: \\ \\ A(7, -2) \\ C(-3, -1) \\ \\ d_{AC}=\sqrt{(7-(-3))^2+(-2-(-1))^2} \\ \\ d_{AC}=\sqrt{(7+3)^2+(-2+1)^2} \\ \\ d_{AC}=\sqrt{(10)^2+(-1)^2} \\ \\ d_{AC}=\sqrt{100+1} \\ \\ \boxed{d_{AC}=\sqrt{101}}[/tex]
[tex]\bullet \ For \ \overline{AC}: \\ \\ B(-4, 9) \\ C(-3, -1) \\ \\ d_{BC}=\sqrt{(-4-(-3))^2+(9-(-1))^2} \\ \\ d_{BC}=\sqrt{(-4+3)^2+(9+1)^2} \\ \\ d_{BC}=\sqrt{(-1)^2+(10)^2} \\ \\ d_{BC}=\sqrt{1+100} \\ \\ \boxed{d_{BC}=\sqrt{101}}[/tex]
Since we have two equal sides, that is:
[tex]d_{AC}=d_{BC}=\sqrt{101}[/tex]
Then the triangle is isosceles.
Scalene triangle: https://brainly.com/question/10379190
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