an isosceles triangle has at least two sides of equal length. Determine whether the triangle with vertices (0,0), (3,4),(7,1) is isosceles.
To determine whether the triangle with vertices (0,0), (3,4), and (7,1) is isosceles, we need to calculate the length of each side of the triangle.
The distance between two points (x1, y1) and (x2, y2) can be calculated using the distance formula:
Distance = sqrt((x2 - x1)^2 + (y2 - y1)^2)
Let's calculate the lengths of the sides using the distance formula:
1. Distance between (0,0) and (3,4):
Distance = sqrt((3-0)^2 + (4-0)^2)
= sqrt(9 + 16)
= sqrt(25)
= 5
2. Distance between (3,4) and (7,1):
Distance = sqrt((7-3)^2 + (1-4)^2)
= sqrt(16 + 9)
= sqrt(25)
= 5
3. Distance between (7,1) and (0, 0):
Distance = sqrt((0-7)^2 + (0-1)^2)
= sqrt(49 + 1)
= sqrt(50)
= 5√2
The lengths of the sides of the triangle are 5, 5, and 5√2. Since all three sides are not equal, the triangle with vertices (0,0), (3,4), and (7,1) is not isosceles.
To determine whether a triangle is isosceles, we need to check if it has at least two sides of equal length. Let's calculate the lengths of the three sides of the given triangle using the distance formula:
Side AB: √((x2-x1)² + (y2-y1)²)
Side BC: √((x3-x2)² + (y3-y2)²)
Side AC: √((x3-x1)² + (y3-y1)²)
Given the vertices:
A (0, 0)
B (3, 4)
C (7, 1)
Calculating the lengths of the sides:
AB = √((3-0)² + (4-0)²)
BC = √((7-3)² + (1-4)²)
AC = √((7-0)² + (1-0)²)
Simplifying the calculations:
AB = √(3² + 4²)
BC = √(4² + (-3)²)
AC = √(7² + 1²)
AB = √(9 + 16) = √25 = 5
BC = √(16 + 9) = √25 = 5
AC = √(49 + 1) = √50 ≈ 7.07
Comparing the lengths:
AB = 5
BC = 5
AC ≈ 7.07
Since the triangle has two sides of equal length (AB = BC = 5), it is an isosceles triangle.
well the distance between the first and second point is 5.
the distance between the first and the last is sqrt50
the distance between the second and the last is sqrt(16+9)=5
so what do you think>