# Concurrency of Three Lines

We will learn how to find the condition of concurrency of three straight lines.

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

Thus, if three lines are concurrent the point of intersection of two lines lies on the third line.

Let the equations of the three concurrent straight lines be

a$$_{1}$$ x + b$$_{1}$$y + c$$_{1}$$  = 0   ……………. (i)

a$$_{2}$$ x + b$$_{2}$$ y + c$$_{2}$$ = 0  ……………. (ii) and

a$$_{3}$$ x + b$$_{3}$$ y + c$$_{3}$$ = 0 ……………. (iii)

Clearly, the point of intersection of the lines (i) and (ii) must be satisfies the third equation.

Suppose the equations (i) and (ii) of two intersecting lines intersect at P(x$$_{1}$$, y$$_{1}$$). Then (x$$_{1}$$, y$$_{1}$$) will satisfy both the equations (i) and (ii).

Therefore, a$$_{1}$$x$$_{1}$$ + b$$_{1}$$y$$_{1}$$  + c$$_{1}$$ = 0 and

a$$_{2}$$x$$_{1}$$ + b$$_{2}$$y$$_{1}$$ + c$$_{2}$$ = 0

Solving the above two equations by using the method of cross-multiplication, we get,

$$\frac{x_{1}}{b_{1}c_{2} - b_{2}c_{1}} = \frac{y_{1}}{c_{1}a_{2} - c_{2}a_{1}} = \frac{1}{a_{1}b_{2} - a_{2}b_{1}}$$

Therefore, x$$_{1}$$  = $$\frac{b_{1}c_{2} - b_{2}c_{1}}{a_{1}b_{2} - a_{2}b_{1}}$$ and

y$$_{1}$$  = $$\frac{c_{1}a_{2} - c_{2}a_{1}}{a_{1}b_{2} - a_{2}b_{1}}$$,  a$$_{1}$$b$$_{2}$$ - a$$_{2}$$b$$_{1}$$ ≠ 0

Therefore, the required co-ordinates of the point of intersection of the lines (i) and (ii) are

($$\frac{b_{1}c_{2} - b_{2}c_{1}}{a_{1}b_{2} - a_{2}b_{1}}$$, $$\frac{c_{1}a_{2} - c_{2}a_{1}}{a_{1}b_{2} - a_{2}b_{1}}$$), a$$_{1}$$b$$_{2}$$ - a$$_{2}$$b$$_{1}$$ ≠ 0

Since the straight lines (i), (ii) and (ii) are concurrent, hence (x$$_{1}$$, y$$_{1}$$) must satisfy the equation (iii).

Therefore,

a$$_{3}$$x$$_{1}$$ + b$$_{3}$$y$$_{1}$$ + c$$_{3}$$ = 0

⇒ a$$_{3}$$($$\frac{b_{1}c_{2} - b_{2}c_{1}}{a_{1}b_{2} - a_{2}b_{1}}$$) + b$$_{3}$$($$\frac{c_{1}a_{2} - c_{2}a_{1}}{a_{1}b_{2} - a_{2}b_{1}}$$) + c$$_{3}$$ = 0

a$$_{3}$$(b$$_{1}$$c$$_{2}$$ - b$$_{2}$$c$$_{1}$$) + b$$_{3}$$(c$$_{1}$$a$$_{2}$$ - c$$_{2}$$a$$_{1}$$) + c$$_{3}$$(a$$_{1}$$b$$_{2}$$ - a$$_{2}$$b$$_{1}$$) = 0

$\begin{vmatrix} a_{1} & b_{1} & c_{1}\\ a_{2} & b_{2} & c_{2}\\ a_{3} & b_{3} & c_{3} \end{vmatrix} = 0$

This is the required condition of concurrence of three straight lines.

Solved example using the condition of concurrency of three given straight lines:

Show that the lines 2x - 3y + 5 = 0, 3x + 4y - 7 = 0 and 9x - 5y + 8 =0 are concurrent.

Solution:

We know that if the equations of three straight lines  a$$_{1}$$ x + b$$_{1}$$y + c$$_{1}$$  = 0, a$$_{2}$$ x + b$$_{2}$$ y + c$$_{2}$$ = 0 and a$$_{3}$$ x + b$$_{3}$$ y + c$$_{3}$$ = 0 are concurrent then

$\begin{vmatrix} a_{1} & b_{1} & c_{1}\\ a_{2} & b_{2} & c_{2}\\ a_{3} & b_{3} & c_{3} \end{vmatrix} = 0$

The given lines are 2x - 3y + 5 = 0, 3x + 4y - 7 = 0 and 9x - 5y + 8 =0

We have

$\begin{vmatrix} 2 & -3 & 5\\ 3 & 4 & -7\\ 9 & -5 & 8\end{vmatrix}$

= 2(32 - 35) - (-3)(24 + 63) + 5(-15 - 36)

= 2(-3) + 3(87) + 5(-51)

= - 6 + 261 -255

= 0

Therefore, the given three straight lines are concurrent.

The Straight Line