Find the equations to the sides of the triangles the coordinates of whose angular points (0, 1), (2, 0) and (-1, -2)

(0, 1), (2, 0) and (-1, -2)

Given:

Points A (0, 1), B (2, 0) and C (-1, -2).

Let us assume,

m_1, m_2, and m₃ be the slope of the sides AB, BC and CA, respectively.

So,

The equation of the line passing through the two points (x₁, y₁) and (x₂, y₂).

Then,

\( m_3=\dfrac{1+2}{1+0}\)

m₁ = \( \dfrac{ -1}{2} \), m₂ = \( \dfrac{ -2}{3} \) and m₃= 3

So, the equation of the sides AB, BC and CA are

By using the formula,

y – y₁= m (x – x₁)

=> y – 1 = (\( \dfrac{ -1}{2} \)) (x – 0)

2y – 2 = -x

x + 2y = 2

=> y – 0 = (\( \dfrac{ -2}{3} \)) (x – 2)

3y = -2x + 4

2x – 3y = 4

= y + 2 = 3(x+1)

y + 2 = 3x + 3

y – 3x = 1

So, we get

x + 2y = 2, 2x – 3y =4 and y – 3x = 1

The equation of sides are x + 2y = 2, 2x – 3y =4 and y – 3x = 1