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





11 and 12 Grade Math 

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