Question

Three lines represented by the equations $a_1x+b_1y+c_1=0$, $a_2x+b_2y+c_2=0$ and $a_3x+b_3y+c_3=0$ are concurrent if;

• A. $a_1(b_2c_3-b_3c_2)+a_2(b_3c_1-b_1c_3)+a_3(b_1c_2-b_2c_1)=0$
• B. $a_1b_2c_3+a_2b_3c_1+a_3b_1c_2=0$
• C. $\frac{a_1}{a_3}=\frac{b_1}{b_3}=\frac{c_1}{c_2}$
• D. $a_1b_2-a_2b_1=0$

Answer 1

Answer: (A)

$a_1(b_2c_3-b_3c_2)+a_2(b_3c_1-b_1c_3)+a_3(b_1c_2-b_2c_1)=0$

Answer 2

For three lines given by the equations:

1. $a_1x+b_1y+c_1=0$
2. $a_2x+b_2y+c_2=0$
3. $a_3x+b_3y+c_3=0$

to be concurrent (i.e., intersect at a single common point), the determinant of their coefficients must be zero.

The condition is:

$$\begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} = 0$$

Expanding this determinant along the first column:

$$a_1 \begin{vmatrix} b_2 & c_2 \\ b_3 & c_3 \end{vmatrix} - a_2 \begin{vmatrix} b_1 & c_1 \\ b_3 & c_3 \end{vmatrix} + a_3 \begin{vmatrix} b_1 & c_1 \\ b_2 & c_2 \end{vmatrix} = 0$$

Calculating the $2 \times 2$ determinants:

$$a_1(b_2c_3 - b_3c_2) - a_2(b_1c_3 - b_3c_1) + a_3(b_1c_2 - b_2c_1) = 0$$

Rearranging the second term:

$$a_1(b_2c_3 - b_3c_2) + a_2(b_3c_1 - b_1c_3) + a_3(b_1c_2 - b_2c_1) = 0$$

This matches the first option.