A member of the family of lines `(x+ y -1) + lambda(2x +3y-2) =0, lambda in R` intersects a member of the family of lines `(ax+y-2) + alpha(bx + 4y-5) = 0, alpha in R` , at a fixed point. The ordered pair (a, b) is

Options

A. (2,5)
B. (1,2)
C. (1,0)
D. none of these

Answer

To solve the problem, we need to find the ordered pair \((a, b)\) such that the lines from two families intersect at a fixed point. Let's break down the solution step by step.

### Step 1: Identify the first family of lines
The first family of lines is given by:
\[ (x + y - 1) + \lambda(2x + 3y - 2) = 0 \]
This can be rewritten as:
\[ L_1 + \lambda L_2 = 0 \]
where \(L_1 = x + y - 1\) and \(L_2 = 2x + 3y - 2\).

### Step 2: Find the intersection point of the first family
To find the fixed point where the lines intersect, we need to solve the equations \(L_1 = 0\) and \(L_2 = 0\):
1. From \(L_1 = 0\):
\[ x + y - 1 = 0 \implies y = 1 - x \]
2. Substitute \(y\) into \(L_2 = 0\):
\[ 2x + 3(1 - x) - 2 = 0 \]
Simplifying this:
\[ 2x + 3 - 3x - 2 = 0 \implies -x + 1 = 0 \implies x = 1 \]
Now substituting \(x = 1\) back into \(y = 1 - x\):
\[ y = 1 - 1 = 0 \]
Thus, the fixed point \(P\) is \((1, 0)\).

### Step 3: Identify the second family of lines
The second family of lines is given by:
\[ (ax + y - 2) + \alpha(bx + 4y - 5) = 0 \]
This can be rewritten as:
\[ L_3 + \alpha L_4 = 0 \]
where \(L_3 = ax + y - 2\) and \(L_4 = bx + 4y - 5\).

### Step 4: Find the conditions for the second family to pass through the fixed point
For the lines from the second family to also pass through the fixed point \(P(1, 0)\), we substitute \(x = 1\) and \(y = 0\) into both equations:

1. For \(L_3 = 0\):
\[ a(1) + 0 - 2 = 0 \implies a - 2 = 0 \implies a = 2 \]

2. For \(L_4 = 0\):
\[ b(1) + 4(0) - 5 = 0 \implies b - 5 = 0 \implies b = 5 \]

### Step 5: Conclusion
The ordered pair \((a, b)\) is:
\[ (a, b) = (2, 5) \]

### Final Answer
The ordered pair \((a, b)\) is \((2, 5)\).