Condition for Concurrency of Three Straight Lines

Let the three straight lines in linear form be \[L_1:A_1x+B_1y+C_1=0\] \[L_2:A_2x+B_2y+C_2=0\] \[L_3:A_3x+B_3y+C_3=0\] Solving equations of $L_1$ and $L_2$, \[\frac{x}{B_1C_2-B_2C_1}=\frac{y}{C_1A_2-A_1C_2}=\frac{1}{A_1B_2-A_2B_1}\] \[\therefore (x,y)=\left(\frac{B_1C_2-B_2C_1}{A_1B_2-A_2B_1},\frac{C_1A_2-C_2A_1}{A_1B_2-A_2B_1}\right)\] is the coordinates of the point of intersection of the lines $L_1$ and $L_2$. Now, the three lines will be concurrent if $L_3$ also passes through the above point of intersection, i.e., \[\text{if }A_3\left(\frac{B_1C_2-B_2C_1}{A_1B_2-A_2B_1}\right)+B_3\left(\frac{C_1A_2-C_2A_1}{A_1B_2-A_2B_1}\right)+C_3=0\] \[A_3B_1C_2-A_3B_2C_1+B_3C_1A_2-B_3C_2A_1+A_1B_2C_3-A_2B_1C_3=0\] \[A_1(B_2C_3-B_3C_2)-A_2(B_1C_3-B_3C_1)+A_3(B_1C_2-B_2C_1)=0\] which is the condition for concurrency of three straight lines.

The condition for the three lines to be concurrent may also be expressed in determinant form as: \[\left|\begin{array}{c} A_1 & A_2 & A_3 \\ B_1 & B_2 & B_3 \\ C_1 & C_2 & C_3 \end{array}\right|=0\] \[\text{or, }\left|\begin{array}{c} A_1 & B_1 & C_1 \\ A_2 & B_2 & C_2 \\ A_3 & B_3 & C_3 \end{array}\right|=0\]

In practice, we may follow the following rule. If three quantities $l$, $m$, $n$ can be determined such that \[l(A_1x+B_1y+C_1)+m(A_2x+B_2y+C_2)\] \[+n(A_3x+B_3y+C_3)=0\] then the three lines are concurrent.


Prove that the straight lines $x+2y=0$, $3x-4y-10=0$ and $5x+3y-7=0$ are concurrent.

Here, \[x+2y=0\text{ __(1)}\] \[3x-4y-10=0\text{ __(2)}\] \[5x+3y-7=0\text{ __(3)}\] Solving $\text{(1)}$ and $\text{(2)}$, \[\begin{array}{c} 3x+6y &=0 \\ -3x+4y+10 &= 0 \\ \hline \\ 10y+10 &= 0 \\ \therefore y &=-1 \end{array}\] From $\text{(1)}$, \[x-2=0\] \[\therefore x=2\] Thus, the point of intersection of the lines $\text{(1)}$ and $\text{(2)}$ is $(2,-1)$. Given three lines will be concurrent if the third line also passes through this point of intersection, i.e. \[\text{if }5(-2)+3(-1)-7=0\] \[10-3-7=0\] \[0=0\] Hence, the given lines are concurrent.


Prove that the straight lines $9x-13y-90=0$, $2x+11y-20=0$ and $7x+y-70=0$ are concurrent.

Here, \[9x-13y-90=0\text{ __(1)}\] \[2x+11y-20=0\text{ __(2)}\] \[7x+6-70=0\text{ __(3)}\] Solving $\text{(1)}$ and $\text{(2)}$, \[\begin{array}{c} 18x-26y-180 &= 0 \\ -18x-99y+180 &= 0 \\ \hline \\ 73y &= 0 \\ \therefore y &= 0\end{array}\] and, from $\text{(1)}$, \[9x-0-90=0\] \[\therefore x=10\] Hence, the point of intersection of the lines $\text{(1)}$ and $\text{(2)}$ is $(10,0)$. Given three lines will be concurrent if \[7×10+0-70=0\] \[0=0\] Thus, given lines are concurrent.


Find the value of $a$ for which the lines $3x+y-2=0$, $ax+2y-3=0$ and $2x-y-3=0$ may be concurrent. Also find the point of concurrence.

Given lines are, \[3x+y-2=0\text{ __(1)}\] \[ax+2y-3=0\text{ __(2)}\] \[2x-y-3=0\text{ __(3)}\] Solving $\text{(1)}$ and $\text{(3)}$, \[\begin{array}{c} 3x+y-2 &=0 \\ 2x-y-3 &= 0 \\ \hline \\ 5x-5 &= 0 \\ \therefore x &= 1 \end{array}\] and, from $\text{(1)}$, \[3x+y-2=0\] \[\therefore y=-1\] Now, the three lines will be concurrent if \[a-2-3=0\] \[\therefore a=5\] And, the point of concurrence is $(1,-1)$.



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