CLASSIFYING THE TYPES OF TRIANGLES WITH COORDINATES
Equilateral Triangle :
An Equilateral Triangle is a triangle in which all the three sides will be equal.
isosceles Triangle :
A triangle with two equal sides is called as isosceles triangle. The angle corresponding to the equal sides will always be equal.
ScaleneTriangle :
In a scalene triangle the length of all the three sides will be different. And also all the three angles will be different.
Using the formula distance between two points, we can find the length of the sides and by comparing side lengths, we can check which type of triangle it is.
d = √(x2 – x1)2 + (y2 – y1)2
Use the distance formula to classify triangle ABC, as either equilateral, isosceles or scalene :
Problem 1 :
A(3, -1), B(1, 8), C(-6, 1)
Solution :
A(3, -1), B(1, 8), C(-6, 1)
d = √(x2 – x1)2 + (y2 – y1)2
Length of side AB :
A(3, -1), B(1, 8)
AB = √(1 – 3)2 + (8 + 1)2
= √((-2)2 + 92)
= √(4 + 81)
= √85
Length of side BC :
B(1, 8), C(-6, 1)
BC = √(-6 – 1)2 + (1 - 8)2
= √((-7)2 + (-7)2)
= √(49 + 49)
= √98
Length of side CA :
C (-6, 1) and A(3, -1)
CA = √(3 + 6)2 + (-1 - 1)2
= √92 + (-2)2)
= √(81 + 4)
= √85
AB = AC
Since two sides are having equal lengths, it is isosceles triangle.
Problem 2 :
A(1, 0), B(3, 1), C(4, 5)
Solution :
A(1, 0), B(3, 1), C(4, 5)
Length of AB :
(x1, y1) = A(1, 0) (x2, y2) = B(3, 1)
d = √(x2 – x1)2 + (y2 – y1)2
AB = √(3 – 1)2 + (1 - 0)2
= √(22 + 12)
= √(4 + 1)
= √5
Length of BC :
(x1, y1) = B(3, 1) (x2, y2) = C(4, 5)
BC = √(4 – 3)2 + (5 - 1)2
= √(12 + 42)
= √(1 + 16)
= √17
Length of CA :
(x1, y1) = C(4, 5) (x2, y2) = A(1, 0)
CA = √(1 - 4)2 + (0 - 5)2
= √(-3)2 + -52)
= √(9 + 25)
= √34
AB ≠ AC ≠ CA
So, it is scalene triangle.
Problem 3 :
A(-1, 0), B(2, -2), C(4, 1)
Solution :
A(-1, 0), B(2, -2), C(4, 1)
Length of AB :
(x1, y1) = A(-1, 0) (x2, y2) = B(2, -2)
d = √(x2 – x1)2 + (y2 – y1)2
AB = √(2 + 1)2 + (-2 - 0)2
= √(32 + (-2)2)
= √(9 + 4)
= √13
Length of BC :
(x1, y1) = B(2, -2) (x2, y2) = C(4, 1)
BC = √(4 – 2)2 + (1 + 2)2
= √(22 + 32)
= √(4 + 9)
= √13
Length of CA :
(x1, y1) = C(4, 1) (x2, y2) = A(-1, 0)
CA = √(-1 - 4)2 + (0 - 1)2
= √(-5)2 + (-1)2)
= √(25 + 1)
= √26
AB = BC
So, it is isosceles triangle.
Problem 4 :
A(√2, 0), B(-√2, 0), C(0, -√5)
Solution :
A(√2, 0), B(-√2, 0), C(0, -√5)
Length of AB :
(x1, y1) = A(√2, 0) (x2, y2) = B(√-2, 0)
d = √(x2 – x1)2 + (y2 – y1)2
AB = √(√-2 – √2)2 + (0 - 0)2
= √(-2√2)2
= 8
Length of BC :
(x1, y1) = B(√-2, 0) (x2, y2) = C(0, -√5)
BC = √(0 – √-2)2 + (-√5 - 0)2
= √(2 + 5)
= √7
Length of CA :
(x1, y1) = C(0, -√5) (x2, y2) = A(√2, 0)
CA = √(√2 - 0)2 + (0 + √5)2
= √(√2)2 + (√5)2)
= √(2 + 5)
= √7
BC = AC
So, it is isosceles triangle.
Problem 5 :
A(√3, 1), B(-√3, 1), C(0, -2)
Solution :
A(√3, 1), B(-√3, 1), C(0, -2)
Length of AB :
(x1, y1) = (√3, 1) (x2, y2) = (-√3, 1)
d = √(x2 – x1)2 + (y2 – y1)2
AB = √(-√3 – √3)2 + (1 - 1)2
= √((-2√3)2 + 02)
= √12
Length of BC :
(x1, y1) = B(-√3, 1) (x2, y2) = C(0, -2)
BC = √(0 – (-√3))2 + (-2 - 1)2
= √(3 + (-3)2)
= √(3 + 9)
= √12
Length of CA :
(x1, y1) = C (0, -2) (x2, y2) = A (√3, 1)
CA = √(√3 - 0)2 + (1 – (-2))2
= √(√3)2 + 32)
= √(3 + 9)
= √12
AB = BC = CA
So, it is equilateral triangle.
Problem 6 :
A(a, b), B(-a, b), C(0, 2)
Solution :
A(a, b), B(-a, b), C(0, 2)
Length of AB :
(x1, y1) = A(a, b) (x2, y2) = B(-a, b)
d = √(x2 – x1)2 + (y2 – y1)2
AB = √(-a – a)2 + (b - b)2
= √(-2a)2
= -2a
Length of BC :
(x1, y1) = (-a, b) (x2, y2) = (0, 2)
BC = √((0 + a)2 + (2 - b)2)
= √(02 + a2 + 2(a)(0)) + ((2)2 + b2 – 2(2)(b))
= √(a2 + 4 + b2 – 4b)
Length of CA :
(x1, y1) = C(0, 2) (x2, y2) = A(a, b)
CA = √(a - 0)2 + (b - 2)2
= √(a2 + b2 - 2(b)(2) + 22)
= √(a2 + b2 - 4b + 4)
BC = AC
So, it is isosceles triangle.
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