Question

determine whether a triangle with the given vertices is a scalene, isosceles, or equilateral triangle. check

(a) ( t(0,2), r(-6,2), i(-3,-5) )
(b) ( a(8,5), b(4,0), c(8,-5) )
(c) ( d(1,0), e(2,-5), f(-3,-5) )

Explanation:

Step1: Use the distance formula

The distance formula between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\)

Step2: Calculate side lengths for part (a)

For \(T(0,2)\) and \(R(-6,2)\):
\(d_{TR}=\sqrt{(-6 - 0)^2+(2 - 2)^2}=\sqrt{(-6)^2+0^2}=6\)
For \(T(0,2)\) and \(I(-3,-5)\):
\(d_{TI}=\sqrt{(-3 - 0)^2+(-5 - 2)^2}=\sqrt{9 + 49}=\sqrt{58}\)
For \(R(-6,2)\) and \(I(-3,-5)\):
\(d_{RI}=\sqrt{(-3+6)^2+(-5 - 2)^2}=\sqrt{9 + 49}=\sqrt{58}\)
Since \(d_{TI}=d_{RI}
eq d_{TR}\), it is an isosceles triangle.

Step3: Calculate side lengths for part (b)

For \(A(8,5)\) and \(B(4,0)\):
\(d_{AB}=\sqrt{(4 - 8)^2+(0 - 5)^2}=\sqrt{16 + 25}=\sqrt{41}\)
For \(A(8,5)\) and \(C(8,-5)\):
\(d_{AC}=\sqrt{(8 - 8)^2+(-5 - 5)^2}=\sqrt{0 + 100}=10\)
For \(B(4,0)\) and \(C(8,-5)\):
\(d_{BC}=\sqrt{(8 - 4)^2+(-5 - 0)^2}=\sqrt{16 + 25}=\sqrt{41}\)
Since \(d_{AB}=d_{BC}
eq d_{AC}\), it is an isosceles triangle.

Step4: Calculate side lengths for part (c)

For \(D(1,0)\) and \(E(2,-5)\):
\(d_{DE}=\sqrt{(2 - 1)^2+(-5 - 0)^2}=\sqrt{1 + 25}=\sqrt{26}\)
For \(D(1,0)\) and \(F(-3,-5)\):
\(d_{DF}=\sqrt{(-3 - 1)^2+(-5 - 0)^2}=\sqrt{16 + 25}=\sqrt{41}\)
For \(E(2,-5)\) and \(F(-3,-5)\):
\(d_{EF}=\sqrt{(-3 - 2)^2+(-5+5)^2}=\sqrt{25 + 0}=5\)
Since \(d_{DE}
eq d_{DF}
eq d_{EF}\), it is a scalene triangle.

Answer:

(a) Isosceles triangle
(b) Isosceles triangle
(c) Scalene triangle