CLASSIFYING TRIANGLES BY COORDINATES
We can classify triangles by its side length. With the given coordinates by finding the distance between two points, we can find the length of the side.
Equilateral triangle :
If length of all sides are equal, then it is equilateral triangle.
Isosceles triangle :
If two sides are having the same length, one side is having different length, then it is called isosceles triangle.
Scalene triangle :
If all sides are having different length, then it is called scalene triangle.
Right triangle :
To check if the triangle is right triangle, we use Pythagorean theorem.
Rule :
Square of hypotenuse is equal to sum of squares of remaining sides.
The vertices of a triangle are given. Classify the triangle as scalene, isosceles, or equilateral.
Problem 1 :
(-5, 0), (0, 6), (5, 0)
Solution :
A(-5, 0), B(0, 6), C(5, 0)
AB = √(x2 - x1)² + (y2 - y1)²
= √(0 + 5)² + (6 - 0)²
= √5² + 6²
= √(25 + 36)
AB = √61
BC = √(x2 - x1)² + (y2 - y1)²
= √(5 - 0)² + (0 - 6)²
= √5² + (-6)²
= √(25 + 36)
BC = √61
AC = √(x2 - x1)² + (y2 - y1)²
= √(5 + 5)² + (0 - 0)²
= √(10)² + (0)
= √(100)
AC = 10
As the two sides are equal it is isosceles triangle.
Problem 2 :
(0, -3), (0, 3), (3, 0)
Solution :
A(0, -3), B(0, 3), C(3, 0)
AB = √(x2 - x1)² + (y2 - y1)²
= √(0 - 0)² + (3 + 3)²
= √(0)² + (6)²
= √(36)
AB = 6
BC = √(x2 - x1)² + (y2 - y1)²
= √(3 - 0)² + (0 - 3)²
= √(3)² + (-3)²
= √(9 + 9)
BC = √18
AC = √(x2 - x1)² + (y2 - y1)²
= √(3 - 0)² + (0 + 3)²
= √(3)² + (3)²
= √9 + 9
CA = √18
As the two
sides are equal it is isosceles triangle.
Problem 3 :
(-2, 5), (1, -1), (4, 6)
Solution :
A(-2, 5), B(1, -1), C(4, 6)
AB = √(x2 - x1)² + (y2 - y1)²
= √(1 + 2)² + (-1 - 5)²
= √(3)² + (-6)²
= √(9 + 36)
AB = √45
BC = √(x2 - x1)² + (y2 - y1)²
= √(4 - 1)² + (6 + 1)²
= √(3)² + (7)²
= √(9 + 49)
BC = √58
AC = √(x2 - x1)² + (y2 - y1)²
= √(4 + 2)² + (6 - 5)²
= √ (6)² + (1)²
= √ 36 + 1
AC = √37
As the three sides are different it is scalene triangle.
Problem 4 :
(1, 4), (4, 1), (7, 4)
Solution :
A(1, 4), B(4, 1), C(7, 4)
AB = √(x2 - x1)² + (y2 - y1)²
= √(4 - 1)² + (1 - 4)²
= √(3)² + (-3)²
= √(9 + 9)
AB = √18
BC = √(x2 - x1)² + (y2 - y1)²
= √(7 - 4)² + (4 - 1)²
= √(3)² + (3)²
= √(9 + 9)
BC = √18
AC = √(x2 - x1)² + (y2 - y1)²
= √(7 - 1)² + (4 - 4)²
= √(6)² + (0)²
AC = √36
Using Pythagorean theorem,
AC2 = AB2 + BC2
(√36)2 = (√18)2 + (√18)2
36 = 18+18
36 = 36
Since it satisfies Pythagorean theorem, it is a right triangle.
Problem 5 :
(-1, -6), (1, 1), (4, -5)
Solution :
A(-1, -6), B(1, 1), C(4, -5)
AB = √(x2 - x1)² + (y2 - y1)²
= √(1 + 1)² + (1 + 6)²
= √(2)² + (7)²
= √(4 + 49)
AB = √53
BC = √(x2 - x1)² + (y2
- y1)²
= √(4 - 1)² + (-5 - 1)²
= √(3)² + (-6)²
= √(9 + 36)
BC = √45
AC = √(x2 - x1)² + (y2
- y1)²
= √(4 + 1)² + (-5 + 6)²
= √(5)² + (1)²
= √25 + 1
CA = √26
As the three sides are different, it is scalene triangle.
Problem 6 :
(-4, 3), (2, -1), (8, -1)
Solution :
A(-4, 3), B(2, -1), C(8, -1)
AB = √(x2 - x1)² + (y2 - y1)²
= √(2 + 4)² + (-1 - 3)²
= √(6)² + (-4)²
= √(36 + 16)
AB = √52
BC = √(x2 - x1)² + (y2 - y1)²
= √(8 - 2)² + (-1 + 1)²
= √6²
= √36
BC = 6
AC = √ (x2 - x1)² + (y2 - y1)²
= √ (8 + 4)² + (-1 - 3)²
= √ (12)² + (-4)²
= √144 + 16
AC = √160
As the three
sides are different it is scalene triangle.
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