Question

Which type of triangle is formed with the points A(1, 7), B(-2, 2), and C(4, 2) as its vertices?

Answer 1

The type of triangle formed by the points A(1, 7), B(-2, 2), and C(4, 2) as its vertices is a scalene triangle.

Step-by-step explanation:

The type of triangle formed by the points A(1, 7), B(-2, 2), and C(4, 2) as its vertices can be determined by examining the lengths of its sides. To do this, we can use the distance formula:

d = sqrt( (x2 - x1)^2 + (y2 - y1)^2 )

Calculating the distances AB, BC, and CA:

AB = sqrt( (-2 - 1)^2 + (2 - 7)^2 ) = sqrt(9 + 25) = sqrt(34)

BC = sqrt( (4 - (-2))^2 + (2 - 2)^2 ) = sqrt(36 + 0) = sqrt(36) = 6

CA = sqrt( (1 - 4)^2 + (7 - 2)^2 ) = sqrt(9 + 25) = sqrt(34)

Since AB, BC, and CA are all different lengths, the triangle formed is a scalene triangle. Scalene triangles have three sides of different lengths.

Answer 2

We will have to use the distance formula in order to determine the lengths of each side of the triangle.

Distance formula:
`\sqrt{(x_(2) - x_(1))^(2) + (y_(2) - y_(1))^(2) }`

Let's calculate AB first:
A (1, 7) and B (-2, 2)
A: x1 = 1 and y1 = 7
B: x2 = -2 and y2 = 2

so

`\sqrt{(-2 - 1)^(2) + (2 - 7)^(2) }`

`\sqrt{(-3)^(2) + (-5)^(2) }`

`√(9 + 25 )`
AB =
`√(34)` or (rounded to the nearest tenth) ≈ 5.8

Now let's do BC:
B: x1 = -2 and y1 = 2
C: x2 = 4 and y2 = 2

So

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

`\sqrt{(6)^(2) + (0)^(2) }`
BC =
`√(36 )` or 6

Now let's do CA
C: x1 = 4 and y1 = 2
A: x2 = 1 and y2 = 7

So

`\sqrt{(1 - 4)^(2) + (7 - 2)^(2) }`

`\sqrt{(-3)^(2) + (5)^(2) }`

`√(9 + 25)`
CA =
`√(34)` or (rounded to the nearest tenth) ≈ 5.8

So let's recap:

AB ≈ 5.8
BC = 6
CA ≈ 5.8

So AB and AC are the same length while BC is .2 units longer which means this is an isosceles triangle.