Final answer:
The distances between the points A(7, -2), B(-4, 9), and C(-3, -1) are found using the distance formula. AB ≅ 15.56 units, AC ≅ 10.05 units, and BC ≅ 10.05 units, making ∆ABC an isosceles triangle.
Step-by-step explanation:
To find the measures of the sides of ∆ABC, we'll begin by using the distance formula, which is √((x2 - x1)^2 + (y2 - y1)^2), to calculate the distances between the points A(7, -2), B(-4, 9), and C(-3, -1).
First, let's find the distance between points A and B:
AB = √((-4 - 7)^2 + (9 - (-2))^2) = √((-11)^2 + (11)^2) = √(121 + 121) = √(242) ≅ 15.56 units
Next, the distance between points A and C:
AC = √((-3 - 7)^2 + (-1 - (-2))^2) = √((-10)^2 + (1)^2) = √(100 + 1) = √(101) ≅ 10.05 units
Lastly, the distance between points B and C:
BC = √((-3 - (-4))^2 + (-1 - 9)^2) = √((1)^2 + (-10)^2) = √(1 + 100) = √(101) ≅ 10.05 units
With these lengths, we can classify ∆ABC by its sides. Since two sides (AC and BC) are equal, ∆ABC is an isosceles triangle.