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

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
$\sqrt{(-2 - 1)^{2} + (2 - 7)^{2} }$
$\sqrt{(-3)^{2} + (-5)^{2} }$
$\sqrt{9 + 25 }$
AB = $\sqrt{34}$ 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 = $\sqrt{36 }$ or 6

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

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