We will learn how to find the condition of concurrency of three straight lines.
Definition of Concurrent Lines:
Three or more lines in a plane are said to be concurrent if all of them
pass through the same point.
In the above Fig., since the three lines ℓ, m and n pass through the point O, these are called concurrent lines.
Also, the point O is called the point of concurrence.
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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