Triangle ABC has vertices A(-3,2) B(4.5) AND C(4,-1) Fin out wether the triangle is isosceles?
find the distances AB, BC, AC are two of them equal?
distance= sqrt((x-x1)^2+(y-y)^2)
It is B’ (6, 5/6
)
To determine whether the triangle ABC is isosceles, we need to compare the lengths of its sides.
Step 1: Calculate the lengths of the three sides of the triangle using the distance formula: d = √((x2 - x1)^2 + (y2 - y1)^2).
- Side AB: d(AB) = √((x2 - x1)^2 + (y2 - y1)^2)
= √((4.5 - (-3))^2 + (0 - 2)^2)
= √((7.5)^2 + (-2)^2)
= √(56.25 + 4)
= √60.25
≈ 7.77
- Side BC: d(BC) = √((x2 - x1)^2 + (y2 - y1)^2)
= √((4 - 4.5)^2 + (-1 - 0)^2)
= √((-0.5)^2 + (-1)^2)
= √(0.25 + 1)
= √1.25
≈ 1.12
- Side AC: d(AC) = √((x2 - x1)^2 + (y2 - y1)^2)
= √((4 - (-3))^2 + (-1 - 2)^2)
= √((7)^2 + (-3)^2)
= √(49 + 9)
= √58
≈ 7.62
Step 2: Compare the lengths of sides AB, BC, and AC.
Since the lengths of sides AB and AC are approximately 7.77 and 7.62 respectively, and the length of side BC is approximately 1.12, we can see that no two sides have the same length. Therefore, the triangle ABC is not isosceles.
Note: While performing the calculations, double-check the coordinates of the given points to ensure accurate results.