Question

Triangle A (1, 1), B (4, 4), C (6, 2) is a: Group of answer choicesIsosceles triangle Right triangle Acute triangle None of the other answers are correct Equilateral triangle

Answer 1

Since
`AB \\eq BC \\eq AC` and
`\cos B = 0`, we conclude that given triangle is a right-angled and scalene.

Explanation:

First, we determine the lengths of sides
`AB`,
`BC` and
`AC` by the Equation of the Line Segment, which is based on the Pythagorean Theorem:

Line segment AB

`AB = \sqrt{(4-1)^(2)+ (4-1)^(2)}`

`AB = 3√(2)`

Line segment BC

`BC = \sqrt{(6-4)^(2)+(2-4)^(2)}`

`BC = 2√(2)`

Line segment AC

`AC = \sqrt{(6-1)^(2)+(2-1)^(2)}`

`AC = √(26)`

And the angles are found by means of the Law of Cosine:, where acute angles exist when
`0 < \cos \theta < 1`, whereas obtuse angles exist for
`-1 < \cos \theta < 0`.

Angle A

`BC^(2) = AB^(2) + AC^(2) - 2\cdot AB\cdot AC \cdot \cos A` (1)

`\cos A = (BC^(2)-AB^(2)-AC^(2))/(-2\cdot AB\cdot AC)`

`\cos A = (8-18-26)/(-2\dot (3√(2))\cdot (√(26)))`

`\cos A = 0.832`

Angle B

`AC^(2) = AB^(2) + BC^(2) - 2\cdot AB\cdot BC\cdot \cos B` (2)

`\cos B = (AC^(2)-AB^(2)-BC^(2))/(-2\cdot AB\cdot BC)`

`\cos B = (26-18-8)/(-2\cdot (3√(2))\cdot (2√(2)))`

`\cos B = 0`

Angle C

`AB^(2) = AC^(2)+BC^(2)-2\cdot AC\cdot BC\cdot \cos C` (3)

`\cos C = (AB^(2)-AC^(2)-BC^(2))/(-2\cdot AC\cdot BC)`

`\cos C = (18 -26-8)/(-2\cdot (√(26))\cdot (2√(2)))`

`\cos C = 0.555`

Since
`AB \\eq BC \\eq AC` and
`\cos B = 0`, we conclude that given triangle is a right-angled and scalene.