Key Concepts and Formulas

• Slope of a Line: For a line given by the equation $Ax + By + C = 0$, the slope $m$ is given by $m = -\frac{A}{B}$.
• Condition for Perpendicularity: Two lines with slopes $m_1$ and $m_2$ are perpendicular if and only if $m_1 \cdot m_2 = -1$.
• Distance Formula: The distance of a point $(x, y)$ from the origin $(0, 0)$ is given by $\sqrt{x^2 + y^2}$.

Step-by-Step Solution

Step 1: Find the slopes of the given lines.

We have two lines: Line 1: $x + (a - 1)y = 1$ Line 2: $2x + a^2y = 1$

Using the formula $m = -\frac{A}{B}$, the slopes are: Slope of Line 1, $m_1 = -\frac{1}{a - 1}$ Slope of Line 2, $m_2 = -\frac{2}{a^2}$

Step 2: Apply the condition for perpendicular lines.

Since the lines are perpendicular, the product of their slopes is -1: $m_1 \cdot m_2 = -1$ $(-\frac{1}{a - 1}) \cdot (-\frac{2}{a^2}) = -1$ $\frac{2}{a^2(a - 1)} = -1$ $2 = -a^2(a - 1)$ $2 = -a^3 + a^2$ $a^3 - a^2 + 2 = 0$

Step 3: Solve for 'a'.

We need to find the value of 'a' that satisfies the equation $a^3 - a^2 + 2 = 0$. By inspection, we can try integer values. Trying $a = -1$, we get: $(-1)^3 - (-1)^2 + 2 = -1 - 1 + 2 = 0$. So, $a = -1$ is a root.

Thus, $(a + 1)$ is a factor. We can perform polynomial division to find the other factor: $(a^3 - a^2 + 2) \div (a + 1) = a^2 - 2a + 2$

So, $a^3 - a^2 + 2 = (a + 1)(a^2 - 2a + 2) = 0$ The quadratic $a^2 - 2a + 2 = 0$ has discriminant $D = (-2)^2 - 4(1)(2) = 4 - 8 = -4 < 0$. Therefore, the quadratic has no real roots. The only real solution is $a = -1$.

Step 4: Find the equations of the lines with a = -1.

Substituting $a = -1$ into the equations of the lines: Line 1: $x + (-1 - 1)y = 1 \Rightarrow x - 2y = 1$ Line 2: $2x + (-1)^2 y = 1 \Rightarrow 2x + y = 1$

Step 5: Find the point of intersection of the two lines.

We have the system of equations: $x - 2y = 1$ (1) $2x + y = 1$ (2)

Multiply equation (2) by 2: $4x + 2y = 2$ (3)

Add equation (1) and (3): $x - 2y + 4x + 2y = 1 + 2$ $5x = 3$ $x = \frac{3}{5}$

Substitute $x = \frac{3}{5}$ into equation (2): $2(\frac{3}{5}) + y = 1$ $\frac{6}{5} + y = 1$ $y = 1 - \frac{6}{5} = -\frac{1}{5}$

So, the point of intersection is $(\frac{3}{5}, -\frac{1}{5})$.

Step 6: Calculate the distance of the point of intersection from the origin.

Using the distance formula, the distance from $(\frac{3}{5}, -\frac{1}{5})$ to $(0, 0)$ is: $d = \sqrt{(\frac{3}{5})^2 + (-\frac{1}{5})^2} = \sqrt{\frac{9}{25} + \frac{1}{25}} = \sqrt{\frac{10}{25}} = \sqrt{\frac{2}{5}}$ $d = \frac{\sqrt{2}}{\sqrt{5}} = \frac{\sqrt{10}}{5}$

However, the correct answer is $\frac{2}{\sqrt{5}}$. Let's re-examine the solution. The error lies in the condition for perpendicularity.

The condition for perpendicularity is correctly applied. The algebra is correct. The point of intersection is correctly calculated. The distance is $\sqrt{\frac{2}{5}}$.

Rationalizing the denominator of the correct answer: $\frac{2}{\sqrt{5}} = \frac{2\sqrt{5}}{5}$ Also, $\sqrt{\frac{2}{5}} = \frac{\sqrt{2}}{\sqrt{5}} = \frac{\sqrt{10}}{5}$

The given correct answer is incorrect. Let's verify that the lines are perpendicular when a = -1. $m_1 = -\frac{1}{-1-1} = \frac{1}{2}$ $m_2 = -\frac{2}{(-1)^2} = -2$ $m_1 \cdot m_2 = \frac{1}{2} \cdot -2 = -1$. The lines are perpendicular.

The derived answer matches option (D).

Common Mistakes & Tips

• Double-check the signs when calculating the slopes of the lines.
• Be careful with algebraic manipulations, especially when substituting values.
• Remember to rationalize the denominator if necessary.

Summary

We found the slopes of the two lines in terms of 'a'. Using the condition for perpendicular lines ($m_1 \cdot m_2 = -1$), we solved for 'a' and found $a = -1$. Substituting this value back into the equations of the lines, we found their equations and solved for their point of intersection. Finally, we calculated the distance of this point from the origin, which is $\sqrt{\frac{2}{5}}$. This result corresponds to option (D). However, the "Correct Answer" provided above is incorrect. The correct answer is actually $\sqrt{\frac{2}{5}}$.

Final Answer

The final answer is $\boxed{\sqrt{2 \over 5}}$, which corresponds to option (D).