Question

given: \\( \triangle a b c, a ( 3,6 ), b ( 5,0 ), c ( 1,0 ) \\)
prove: \\( \triangle a b c \\) is an isosceles triangle.

• \\( c a = \sqrt { ( x _ { 2 } - x _ { 1 } ) ^ { 2 } + ( y _ { 2 } - y _ { 1 } ) ^ { 2 } } \\) so \\( c a = \\)
• \\( a b = \sqrt { ( x _ { 2 } - x _ { 1 } ) ^ { 2 } + ( y _ { 2 } - y _ { 1 } ) ^ { 2 } } \\) so \\( a b = \\)
• \\( \overline { c a } \cong \overline { a b } \\) by definition of
• \\( \triangle a b c \\) is an isosceles triangle by definition of

Explanation:

Step1: Calculate the length of \(CA\)

Use the distance formula \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\). For points \(C(1,0)\) and \(A(3,6)\), \(x_1 = 1,y_1 = 0,x_2=3,y_2 = 6\).

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Step2: Calculate the length of \(AB\)

For points \(A(3,6)\) and \(B(5,0)\), \(x_1 = 3,y_1 = 6,x_2 = 5,y_2=0\).

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Step3: Determine the relationship between \(CA\) and \(AB\)

Since \(CA=\sqrt{40}\) and \(AB = \sqrt{40}\), by the definition of congruent segments (segments with equal length), \(\overline{CA}\cong\overline{AB}\).

Step4: Conclude the type of \(\triangle ABC\)

By the definition of an isosceles triangle (a triangle with at least two congruent sides), \(\triangle ABC\) is an isosceles triangle.

Answer:

• \(CA=\sqrt{(3 - 1)^2+(6 - 0)^2}=2\sqrt{10}\)
• \(AB=\sqrt{(5 - 3)^2+(0 - 6)^2}=2\sqrt{10}\)
• \(\overline{CA}\cong\overline{AB}\) by the definition of congruent segments (segments with equal length)
• \(\triangle ABC\) is an isosceles triangle by the definition of an isosceles triangle (a triangle with at least two congruent sides)