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 thempass 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 bea\(_{1}\) x + b\(_{1}\)y + c\(_{1}\) = 0 ……………. (i)a\(_{2}\) x + b\(_{2}\) y + c\(_{2}\) = 0 ……………. (ii) anda\(_{3}\) 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 anda\(_{2}\)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\(_{1}\) = \(\frac{b_{1}c_{2} –
b_{2}c_{1}}{a_{1}b_{2} – a_{2}b_{1}}\) andy\(_{1}\) = \(\frac{c_{1}a_{2} – c_{2}a_{1}}{a_{1}b_{2} –
a_{2}b_{1}}\), a\(_{1}\)b\(_{2}\) – a\(_{2}\)b\(_{1}\) ≠ 0Therefore, 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\(_{1}\) ≠ 0Since 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 =0We 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= 0Therefore, the given three straight lines are concurrent.
● The Straight Line
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