Question 4
Check whether (5, –2), (6, 4) and (7, –2) are the vertices of an isosceles triangle.
📋 Given Information
- First vertex: \( A(5, -2) \) where \( x_1 = 5, y_1 = -2 \)
- Second vertex: \( B(6, 4) \) where \( x_2 = 6, y_2 = 4 \)
- Third vertex: \( C(7, -2) \) where \( x_3 = 7, y_3 = -2 \)
🎯 To Find
Whether the three given points form an isosceles triangle.
Definition: An isosceles triangle is a triangle that has at least two sides of equal length.
Types of Triangles by Sides:
- Scalene Triangle: All three sides are different
- Isosceles Triangle: Two sides are equal
- Equilateral Triangle: All three sides are equal
Method: Calculate all three sides using the distance formula. If any two sides are equal, it’s an isosceles triangle.
📐 Formula
📝 Step-by-Step Solution
Using points A(5, -2) and B(6, 4):
\[ AB = \sqrt{(6 – 5)^2 + (4 – (-2))^2} \] \[ AB = \sqrt{(1)^2 + (6)^2} \] \[ AB = \sqrt{1 + 36} = \sqrt{37} \text{ units} \]Using points B(6, 4) and C(7, -2):
\[ BC = \sqrt{(7 – 6)^2 + (-2 – 4)^2} \] \[ BC = \sqrt{(1)^2 + (-6)^2} \] \[ BC = \sqrt{1 + 36} = \sqrt{37} \text{ units} \]Using points A(5, -2) and C(7, -2):
\[ AC = \sqrt{(7 – 5)^2 + (-2 – (-2))^2} \] \[ AC = \sqrt{(2)^2 + (0)^2} \] \[ AC = \sqrt{4 + 0} = \sqrt{4} = 2 \text{ units} \]We have:
- \( AB = \sqrt{37} \) units
- \( BC = \sqrt{37} \) units
- \( AC = 2 \) units
Observation: \( AB = BC = \sqrt{37} \)
Since two sides are equal, the triangle is isosceles.
🔍 Additional Observations
- Equal sides: AB = BC = √37 ≈ 6.08 units
- Base: AC = 2 units
- Type: Isosceles triangle (two equal sides)
- Special property: Points A and C have the same y-coordinate (-2), so AC is horizontal
- Symmetry: Point B is equidistant from A and C
📊 Verification: Check if it’s also a Right Triangle
Let’s check if this triangle satisfies the Pythagorean theorem:
For a right triangle: \( a^2 + b^2 = c^2 \) (where c is the longest side)
We have: AB = BC = √37, AC = 2
Check: \( AC^2 + BC^2 = AB^2 \)?
\[ 2^2 + (\sqrt{37})^2 = 4 + 37 = 41 \] \[ AB^2 = (\sqrt{37})^2 = 37 \]Since \( 41 \neq 37 \), this is NOT a right triangle.
Alternatively, check: \( AC^2 + AB^2 = BC^2 \)?
\[ 4 + 37 = 41 \neq 37 \]Conclusion: It’s an isosceles triangle but NOT a right triangle.
⚠️ Common Mistakes
❌ Mistake 1: Calculating only two sides
✓ Correct: Always calculate all three sides to properly compare
❌ Mistake 2: Sign error with negative coordinates
Writing \( 4 – (-2) = 2 \) instead of 6
✓ Correct: \( 4 – (-2) = 4 + 2 = 6 \)
❌ Mistake 3: Forgetting to simplify √4
Leaving AC = √4 instead of 2
✓ Correct: \( \sqrt{4} = 2 \)
❌ Mistake 4: Confusing isosceles with equilateral
Isosceles = 2 equal sides, Equilateral = 3 equal sides
✓ Correct: Check all three sides before concluding
💡 Key Points to Remember
- Isosceles Triangle: At least two sides are equal
- Equilateral Triangle: All three sides are equal
- Scalene Triangle: All three sides are different
- Distance Formula: \( \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \)
- Comparison: Calculate all sides before making conclusions
- Special case: If two points share same x or y coordinate, distance is simpler
- Verification: Check if Pythagorean theorem holds for right triangle
📝 Practice Similar Problems
- Question 5: Check if quadrilateral is a square
- Question 6: Find fourth vertex of parallelogram
- Try: Check if (0, 0), (3, 4), (6, 0) form an isosceles triangle
