Triangle def has vertices d(2,-5) e(6,-1) and f(5,6) verify that def is isosceles
1 answer:
To prove that the given triangle is an isosceles, two of its sides must have equal lengths.
Between D(2,-5) and E(6,-1):
d = √(6 - 2)² + (-1 - ⁻5)² = 4√2
Between E(6,-1) and F(5,6):
d = √(5 - 6)² + (6 - ⁻1)² = 5√2
Between D(2,-5) and F(5,6)
d = √(5 - 2)² + (6 - ⁻5)² = √130
<em>Since there are no equal lengths, the triangle is NOT an isosceles triangle.</em>
The bisector of angle F is a line segment from vertex F to the midpoint of line DE.
Midpoint: x = (2+6)/2 = 4; y = (-5+-1)/2 = -3
The midpoint is at (4,-3).
The equation of the line follows the point-slope formula:
y - y₁ = [(y₂ - y₁)/(x₂ - x₁)](x - x₁)
Substituting using points (5,6) and (4,-3)
y - 6 = [(⁻3 - 6)/(4 - 6)](x - 5)
y - 6 = (9/2)(x - 5)
2(y - 6) = 9(x - 5)
2y - 12 = 9x - 45
<em>2y - 9x = -33 </em>
Between D(2,-5) and E(6,-1):
d = √(6 - 2)² + (-1 - ⁻5)² = 4√2
Between E(6,-1) and F(5,6):
d = √(5 - 6)² + (6 - ⁻1)² = 5√2
Between D(2,-5) and F(5,6)
d = √(5 - 2)² + (6 - ⁻5)² = √130
<em>Since there are no equal lengths, the triangle is NOT an isosceles triangle.</em>
The bisector of angle F is a line segment from vertex F to the midpoint of line DE.
Midpoint: x = (2+6)/2 = 4; y = (-5+-1)/2 = -3
The midpoint is at (4,-3).
The equation of the line follows the point-slope formula:
y - y₁ = [(y₂ - y₁)/(x₂ - x₁)](x - x₁)
Substituting using points (5,6) and (4,-3)
y - 6 = [(⁻3 - 6)/(4 - 6)](x - 5)
y - 6 = (9/2)(x - 5)
2(y - 6) = 9(x - 5)
2y - 12 = 9x - 45
<em>2y - 9x = -33 </em>
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