If the lines `x+(a-1)y=1` and `2x+1a^(2)y=1` there `ainR-{0,1}` are perpendicular to each other, Then distance of their point of intersection from the origin is

Options

A. `sqrt(2/5)`
B. `2/5`
C. `2/(sqrt(5))`
D. `(sqrt(2))/5`

Answer

To solve the problem of finding the distance of the point of intersection of the lines \(x + (a-1)y = 1\) and \(2x + a^2y = 1\) from the origin, given that these lines are perpendicular, we can follow these steps:

### Step 1: Rewrite the equations of the lines in slope-intercept form
The given equations are:
1. \(x + (a-1)y = 1\)
2. \(2x + a^2y = 1\)

We can rearrange these equations to find the slopes.

For the first line:
\[ (a-1)y = 1 - x \implies y = \frac{1 - x}{a-1} = -\frac{1}{a-1}x + \frac{1}{a-1} \]
Thus, the slope \(m_1\) of the first line is:
\[ m_1 = -\frac{1}{a-1} \]

For the second line:
\[ a^2y = 1 - 2x \implies y = \frac{1 - 2x}{a^2} = -\frac{2}{a^2}x + \frac{1}{a^2} \]
Thus, the slope \(m_2\) of the second line is:
\[ m_2 = -\frac{2}{a^2} \]

### Step 2: Use the condition for perpendicular lines
For the lines to be perpendicular, the product of their slopes must equal \(-1\):
\[ m_1 \cdot m_2 = -1 \]
Substituting the slopes:
\[ \left(-\frac{1}{a-1}\right) \cdot \left(-\frac{2}{a^2}\right) = -1 \]
This simplifies to:
\[ \frac{2}{a^2(a-1)} = -1 \]
Multiplying both sides by \(-a^2(a-1)\):
\[ 2 = -a^2(a-1) \]
This leads to:
\[ a^3 - a^2 + 2 = 0 \]

### Step 3: Find the roots of the cubic equation
We can use the Rational Root Theorem or trial and error to find roots. Testing \(a = -1\):
\[ (-1)^3 - (-1)^2 + 2 = -1 - 1 + 2 = 0 \]
Thus, \(a = -1\) is a root.

### Step 4: Factor the cubic equation
Using synthetic division or polynomial long division, we can factor the cubic equation:
\[ a^3 - a^2 + 2 = (a + 1)(a^2 - 2a + 2) \]
The quadratic \(a^2 - 2a + 2\) has a discriminant \(D = (-2)^2 - 4 \cdot 1 \cdot 2 = 4 - 8 = -4\), which means it has no real roots. Thus, the only real solution is \(a = -1\).

### Step 5: Substitute \(a = -1\) back into the line equations
Substituting \(a = -1\) into the original line equations:
1. \(x + (-1-1)y = 1 \implies x - 2y = 1\)
2. \(2x + (-1)^2y = 1 \implies 2x + y = 1\)

### Step 6: Solve the system of equations
We can solve the system:
1. \(x - 2y = 1\) (Equation 1)
2. \(2x + y = 1\) (Equation 2)

From Equation 1, we can express \(x\) in terms of \(y\):
\[ x = 1 + 2y \]
Substituting into Equation 2:
\[ 2(1 + 2y) + y = 1 \implies 2 + 4y + y = 1 \implies 5y = -1 \implies y = -\frac{1}{5} \]
Substituting \(y\) back to find \(x\):
\[ x = 1 + 2\left(-\frac{1}{5}\right) = 1 - \frac{2}{5} = \frac{3}{5} \]
Thus, the point of intersection is \(\left(\frac{3}{5}, -\frac{1}{5}\right)\).

### Step 7: Calculate the distance from the origin
The distance \(d\) from the origin \((0, 0)\) to the point \(\left(\frac{3}{5}, -\frac{1}{5}\right)\) is given by:
\[ d = \sqrt{\left(\frac{3}{5} - 0\right)^2 + \left(-\frac{1}{5} - 0\right)^2} = \sqrt{\left(\frac{3}{5}\right)^2 + \left(-\frac{1}{5}\right)^2} \]
Calculating:
\[ d = \sqrt{\frac{9}{25} + \frac{1}{25}} = \sqrt{\frac{10}{25}} = \sqrt{\frac{2}{5}} \]

### Final Answer
The distance of the point of intersection from the origin is \(\sqrt{\frac{2}{5}}\).