Question

Prove that the points (5, -2), (6, 4) and (7, -2) are the vertices of an isosceles triangle.

Hint:

When verifying a triangle’s type, use the distance formula and compare the side lengths to check for equality.

Answer 1

Step 1: Use distance formula.
Distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \] Step 2: Calculate sides.
\[ AB = \sqrt{(6 - 5)^2 + (4 - (-2))^2} = \sqrt{1^2 + 6^2} = \sqrt{37} \] \[ BC = \sqrt{(7 - 6)^2 + (-2 - 4)^2} = \sqrt{1 + 36} = \sqrt{37} \] \[ CA = \sqrt{(7 - 5)^2 + (-2 - (-2))^2} = \sqrt{4 + 0} = 2 \] Step 3: Compare sides.
AB = BC = $\sqrt{37}$, hence the triangle is isosceles.
Step 4: Conclusion.
Therefore, the points (5, -2), (6, 4), and (7, -2) form an isosceles triangle.