Check whether (5, – 2), (6, 4) and (7, – 2) are the vertices of an isosceles triangle.
Step-by-Step Solution
Key Concept: Use the distance formula to find the lengths of the three sides. In a triangle, if any two sides are equal, the triangle is isosceles. Also verify that the points are non‑collinear (area ≠ 0).
Let the three points be \(A(5,-2),\; B(6,4),\; C(7,-2)\).
1. Find the length of AB using the distance formula:
$$AB = \sqrt{(6-5)^2 + (4-(-2))^2}
= \sqrt{1^2 + 6^2}
= \sqrt{1+36}
= \sqrt{37}.$$
2. Find the length of BC:
$$BC = \sqrt{(7-6)^2 + (-2-4)^2}
= \sqrt{1^2 + (-6)^2}
= \sqrt{1+36}
= \sqrt{37}.$$
3. Find the length of AC:
$$AC = \sqrt{(7-5)^2 + (-2-(-2))^2}
= \sqrt{2^2 + 0^2}
= \sqrt{4}
= 2.$$
4. Compare the side lengths:
\(AB = \sqrt{37}\) and \(BC = \sqrt{37}\) are equal, while \(AC = 2\) is different.
Hence two sides are equal ⇒ the triangle is isosceles.
5. Check that the points are not collinear (optional but recommended):
The slope of AB is \(\frac{4-(-2)}{6-5}=6\), whereas the slope of AC is \(\frac{-2-(-2)}{7-5}=0\). Since the slopes are different, the points are non‑collinear.
Therefore, the points \((5,-2), (6,4), (7,-2)\) indeed form an isosceles triangle with the equal sides \(AB\) and \(BC\).
Correct Answer: Yes, the three points form an isosceles triangle because AB = BC = \(\sqrt{37}\) while AC = 2.