Straight Lines

# 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 $(1,-1)$.