CLASSIFYING THE TYPES OF TRIANGLES WITH COORDINATES

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.