If the lines `a_1x+b_1y+1=0,\ a_2x+b_2y+1=0\ a n d\ a_3x+b_3y+1=0` are concurrent, show that the points `(a_1, b_1),\ (a_2, b_2)a n d\ (a_3, b_3)` are collinear.

Explanation

To show that the points \((a_1, b_1)\), \((a_2, b_2)\), and \((a_3, b_3)\) are collinear if the lines \(a_1x + b_1y + 1 = 0\), \(a_2x + b_2y + 1 = 0\), and \(a_3x + b_3y + 1 = 0\) are concurrent, we can follow these steps:

### Step 1: Understand the condition for concurrency
The lines are said to be concurrent if they intersect at a single point. This means that the determinant formed by the coefficients of the lines must be equal to zero.

### Step 2: Set up the determinant
The determinant for the points \((a_1, b_1)\), \((a_2, b_2)\), and \((a_3, b_3)\) can be set up as follows:

...