Is the triangle with vertices D (-1: -2) E (1: 0) F (2: -3) isosceles?

To find out if a triangle is isosceles, you need to find two sides in this triangle with equal length.

To do this, we use the formula for finding the distance between two points with the given coordinates (x1, y1) and (x2, y2):
√ ((x1 – x2) ^ 2 + (y1 – y2) ^ 2) = l is the distance between two points.

Then the distance between points D (-1: -2) and E (1: 0) will be equal to:
√ ((1 – (-1)) ^ 2 + (0 – (-2)) ^ 2) = √ ((1 + 1)) ^ 2 + (0 + 2)) ^ 2)

√ ((1 + 1) ^ 2 + (0 + 2) ^ 2) = √ ((2) ^ 2 + (2) ^ 2) = √ (4 + 4) = √8 = 2√2

Then the distance between points D (-1: -2) and F (2: -3) will be equal to:
√ ((- 1 – 2) ^ 2 + (-2 – (-3)) ^ 2) = √ ((- 3) ^ 2 + (-2 + 3) ^ 2) = √ (3 ^ 2 + 1 ^ 2) = √ (9 + 1) = √10

Then the distance between points E (1: 0) and F (2: -3) will be equal to:
√ ((1 – 2) ^ 2 + (0 – (-3)) ^ 2) = √ ((- 1) ^ 2 + (0 + 3) ^ 2) = √ (1 + 3 ^ 2) = √ (1 + 9) = √10

Answer: since the lengths of the sides EF and DF are equal to √10, then this triangle is isosceles.