Bilello985
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Which type of triangle is formed with the points A(1, 7), B(-2, 2), and C(4, 2) as its vertices?

Sagot :

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

Distance formula: [tex] \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} } [/tex]

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
[tex] \sqrt{(-2 - 1)^{2} + (2 - 7)^{2} } [/tex]
[tex] \sqrt{(-3)^{2} + (-5)^{2} } [/tex]
[tex] \sqrt{9 + 25 } [/tex]
AB = [tex] \sqrt{34} [/tex] 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
[tex]\sqrt{(4 - -2)^{2} + (2 - 2)^{2} }[/tex]
[tex]\sqrt{(6)^{2} + (0)^{2} }[/tex]
BC = [tex]\sqrt{36 }[/tex] or 6

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

So
[tex] \sqrt{(1 - 4)^{2} + (7 - 2)^{2} } [/tex]
[tex] \sqrt{(-3)^{2} + (5)^{2} } [/tex]
[tex] \sqrt{9 + 25}[/tex]
CA = [tex]\sqrt{34}[/tex] 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.

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