Question

Using the Distance formula, Classify the Following Triangle. A(7, -2) B(-4, 9) C(-3,-1)

Answer 1

The triangle is isosceles.

Explanation:

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:

$d=\sqrt{(x_{1}-x_{2})^2+(y_{1}-y_{2})^2}$

Then:

$\bullet \ For \ \overline{AB}: \\ \\ A(7, -2) \\ B(-4, 9) \\ \\ d_{AB}=\sqrt{(7-(-4))^2+(-2-9)^2} \\ \\ d_{AB}=\sqrt{(7+4)^2+(-2-9)^2} \\ \\ d_{AB}=\sqrt{(11)^2+(-11)^2} \\ \\ \boxed{d_{AB}=11\sqrt{2}}$

$\bullet \ For \ \overline{AC}: \\ \\ A(7, -2) \\ C(-3, -1) \\ \\ d_{AC}=\sqrt{(7-(-3))^2+(-2-(-1))^2} \\ \\ d_{AC}=\sqrt{(7+3)^2+(-2+1)^2} \\ \\ d_{AC}=\sqrt{(10)^2+(-1)^2} \\ \\ d_{AC}=\sqrt{100+1} \\ \\ \boxed{d_{AC}=\sqrt{101}}$

$\bullet \ For \ \overline{AC}: \\ \\ B(-4, 9) \\ C(-3, -1) \\ \\ d_{BC}=\sqrt{(-4-(-3))^2+(9-(-1))^2} \\ \\ d_{BC}=\sqrt{(-4+3)^2+(9+1)^2} \\ \\ d_{BC}=\sqrt{(-1)^2+(10)^2} \\ \\ d_{BC}=\sqrt{1+100} \\ \\ \boxed{d_{BC}=\sqrt{101}}$

Since we have two equal sides, that is:

$d_{AC}=d_{BC}=\sqrt{101}$

Then the triangle is isosceles.

Learn more:

Scalene triangle: brainly.com/question/10379190

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