# Conditions of Concurrency of Three Lines

Conditions of Concurrency of Three Lines

Three lines are said to be concurrent if they pass through a common point, i.e., they meet at a point.

Thus, if three lines are concurrent the point of intersection of two lies on the third line. Let a₁x + b₁y + c₁ = 0 … 1

a₂x + b₂y + c₂ = 0 … 2

a₃x + b₃y + c₃ = 0 … 3

Be three concurrent lines. Then the point of intersection of (1) and (2) must lie on the third.

The coordinates of the point of intersection of (1) and (2) are $$\left[ \frac{{{b}_{1}}{{c}_{2}}-{{b}_{2}}{{c}_{1}}}{{{a}_{1}}{{b}_{2}}-{{b}_{1}}{{a}_{2}}},\frac{{{c}_{1}}{{a}_{2}}-{{a}_{1}}{{c}_{2}}}{{{a}_{1}}{{b}_{2}}-{{b}_{1}}{{a}_{2}}} \right]$$.

This point lies on (3), therefore $${{a}_{3}}\left[ \frac{{{b}_{1}}{{c}_{2}}-{{b}_{2}}{{c}_{1}}}{{{a}_{1}}{{b}_{2}}-{{b}_{1}}{{a}_{2}}} \right]+{{b}_{3}}\left[ \frac{{{c}_{1}}{{a}_{2}}-{{a}_{1}}{{c}_{2}}}{{{a}_{1}}{{b}_{2}}-{{b}_{1}}{{a}_{2}}} \right]+{{c}_{3}}=0$$,

⇒ a₃ (b₁c₂ – c₁b₂) + b₃ (c₁a₂ – c₂a₁) + c₃ (a₁b₂ – b₁a₂) = 0

$$\Rightarrow \left| \begin{matrix} {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\ {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\\end{matrix} \right|=0$$.

This is the required condition of concurrency of three lines.

Example: If the lines ax + y + 1 = 0, x + by + 1 = 0 and x + y + c = 0are concurrent (a ≠ b ≠ c ≠ 1) then find the value of  $$\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}$$.

Solution: Since the given lines are concurrent. Therefore, $$\left| \begin{matrix} a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c \\\end{matrix} \right|=0$$,

[Applying C₂ → C₂ – C₁ , C₃ → C₃ – C₁]

= a (b – 1) (c – 1) – (1 – a) (c – 1) – (1 – a) (b – 1) = 0

$$\Rightarrow \frac{a}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=0$$,

$$\Rightarrow \frac{a-1+1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=0$$,

$$\Rightarrow \frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=1$$.