Question

If three lines whose equations are y = m₁x + c₁, y = m₂x + c₂ and y = m₃x + c₃ are concurrent, then show that m₁(c₂ – c₃) + m₂ (c₃ – c1) + m₃ (c₁ – c₂) = 0.

Answer 1

The equations of the given lines are

y = m₁x + c₁   …(1)

y = m₂x + c₂   …(2)

y = m₃x + c₃   …(3)

On subtracting equation (1) from (2), we obtain

0 = (m₂ − m₁)x + (c₂ − c₁₎

= (m₁ − m₂)x = c₂ − c₁

= `x = (c_2 - c_1)/(m_1 - m_2)`

On substituting this value of x in (1), we obtain

`y = m_1 ((c_2 - c_1)/(m_1 - m_2)) + c_1`

`y = ((m_1c_2 - m_1c_1)/(m_1 - m_2)) + c_1`

`y = (m_1c_2 - m_1c_1 + m_1c_1 - m_2c_1)/(m_1 - m_2)`

`y = (m_1c_2 - m_2c_1)/(m_1 - m_2)`

∴ `((c_2 - c_1)/(m_1 - m_2), (m_1c_2 - m_2c_1)/(m_1 - m_2))` is the point of intersection of lines (1) and (2).

It is given that lines (1), (2), and (3) are concurrent. Hence, the point of intersection of lines (1) and (2) will also satisfy equation (3).

= `(m_1c_2 - m_2c_1)/(m_1 - m_2) = m_3 ((c_2 - c_1)/(m_1 - m_1)) + c_3`

= `(m_1c_2 - m_2c_1)/(m_1 - m_2) = (m_3c_2 - m_3c_1 + c_3m_1 - c_3m_2)/(m_1 - m_2)`

= m₁c₂ - m₂c₁ - m₃c₂ + m₃c₁ - c₃m₁ + c₃m₂ = 0

= m₁ (c₂ - c₃) + m₂ (c₃ - c₁) + m₃ (c₁ - c₂) = 0

Hence, m₁ (c₂ - c₃) + m₂ (c₃ - c₁) + m₃ (c₁ - c₂) = 0