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
From Concurrency of Three Lines to HOME PAGE
Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.
Dec 04, 24 01:30 AM
Dec 04, 24 01:07 AM
Dec 04, 24 12:45 AM
Dec 04, 24 12:06 AM
Dec 03, 24 11:37 PM
New! Comments
Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.