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Step-by-Step Solution
Step 1: Write down the given lines
Line 1: $x + (a - 1)\,y = 1$
Line 2: $2x + a^2\,y = 1$
Step 2: Determine the slopes of the two lines
• For Line 1: $x + (a-1)\,y = 1 \implies (a - 1)\,y = 1 - x \implies y = \frac{1 - x}{a - 1}$.
Hence, its slope $m_1 = -\frac{1}{a-1}.$
• For Line 2: $2x + a^2\,y = 1 \implies a^2\,y = 1 - 2x \implies y = \frac{1 - 2x}{a^2}.$
Hence, its slope $m_2 = -\frac{2}{a^2}.$
Step 3: Use the condition for perpendicularity
For two lines to be perpendicular, the product of their slopes is $-1$. Therefore,
$m_1 \cdot m_2 = -1.$
Substituting $m_1$ and $m_2$:
$\left( -\frac{1}{a-1} \right)\left( -\frac{2}{a^2} \right) = -1.$
Step 4: Solve for $a$
Simplify the above equation:
$\frac{2}{(a-1)a^2} = -1$
$\frac{2}{a^3 - a^2} = -1$
Multiply both sides by $a^3 - a^2$ to get:
$2 = -\left(a^3 - a^2\right)$
$a^3 - a^2 + 2 = 0.$
Factorize if possible:
$a^3 - a^2 + 2 = 0.$
Let's factor out $(a + 1)$:
$(a + 1)\,\bigl(a^2 - 2a + 2\bigr) = 0.$
Hence, $a = -1$ (since $a^2 - 2a + 2 = 0$ has no real roots, as its discriminant is negative).
Step 5: Substitute $a = -1$ into the original lines
• Substituting $a = -1$ in Line 1: $x + ( -1 - 1 )\,y = 1 \implies x - 2\,y = 1,$ or
$x - 2y + (-1) = 0 \implies x - 2y + 1 = 0.$
• Substituting $a = -1$ in Line 2: $2x + (-1)^2\,y = 1 \implies 2x + 1\cdot y = 1,$
or $2x + y - 1 = 0.$
Step 6: Find the point of intersection
We solve the following system simultaneously:
1) $x - 2y + 1 = 0,$
2) $2x + y - 1 = 0.$
From the first equation, $x = 2y - 1.$ Substitute this into the second equation:
$2(2y - 1) + y - 1 = 0,$
$4y - 2 + y - 1 = 0,$
$5y - 3 = 0,$
$y = \frac{3}{5}.$
Then, $x = 2\left(\frac{3}{5}\right) - 1 = \frac{6}{5} - 1 = \frac{1}{5}.$
However, from the provided solution steps, it appears their intersection is $\left(\frac{3}{5}, -\frac{1}{5}\right).$ Let us verify carefully:
• Check with the first equation $x - 2y + 1 = 0.$
If $x = \frac{3}{5},\, y = -\frac{1}{5},$
LHS $= \frac{3}{5} - 2\left(-\frac{1}{5}\right) + 1
= \frac{3}{5} + \frac{2}{5} + 1
= \frac{5}{5} + 1
= 2,$ which is not 0.
Thus, we need to redo the substitution correctly to match the lines shown in the final step.
Observe the solution provided:
Line 1 is written as: $x - 2y + 1 = 0,$ which rearranges to $x - 2y = -1,$ or $x = -1 + 2y.$
But originally, from $x + (a-1)y = 1$ and substituting $a=-1,$ we get $x - 2y = 1,$ which rearranges to $x = 1 + 2y.$
So perhaps the correct form is $x - 2y - 1 = 0,$ not $+1 = 0.$
Similarly for Line 2: $2x + y = 1,$ so $2x + y - 1 = 0$ is consistent.
Let’s solve them with consistent forms:
Line 1: $x - 2y = 1 \implies x = 1 + 2y.$
Line 2: $2x + y = 1.$
Substitute $x = 1 + 2y$ into $2x + y = 1:$
$2(1 + 2y) + y = 1,$
$2 + 4y + y = 1,$
$5y + 2 = 1,$
$5y = -1,$
$y = -\frac{1}{5}.$
Then $x = 1 + 2\left(-\frac{1}{5}\right) = 1 - \frac{2}{5} = \frac{5}{5} - \frac{2}{5} = \frac{3}{5}.$
Hence the correct intersection is indeed:
Point $P\left(\frac{3}{5},\, -\frac{1}{5}\right).$
Step 7: Compute the distance of $P$ from the origin
Use the distance formula $OP = \sqrt{x^2 + y^2}:$
$OP = \sqrt{\left(\frac{3}{5}\right)^2 + \left(-\frac{1}{5}\right)^2}
= \sqrt{\frac{9}{25} + \frac{1}{25}}
= \sqrt{\frac{10}{25}}
= \sqrt{\frac{2}{5}}.$
Step 8: State the final answer
The distance of the point of intersection from the origin is
$\sqrt{\frac{2}{5}}.$
