If the two lines x + (a - 1)y = 1 and 2x + a²y = 1 (a ∈ R - {0, 1}) are perpendicular, then the dist

Q: Why do we use $m_1 m_2 = -1$ instead of just solving for the intersection point directly?

If you try to solve for the intersection point first, you will be carrying the variable $a$ through a messy algebraic nightmare. By using the perpendicularity condition first, we 'freeze' the parameter $a$ into a concrete value, turning a variable-heavy problem into a simple, solvable system of linear equations. Always look for the constraint that simplifies the variables first!

Q: How can I be sure $a = -1$ is the only real root for the cubic equation $a^3 - a^2 + 2 = 0$?

That is a fantastic observation. When you factor out $(a+1)$, you are left with a quadratic expression: $a^2 - 2a + 2$. If you check its discriminant ($D = b^2 - 4ac$), you get $4 - 8 = -4$. Since the discriminant is negative, this quadratic has no real roots. Thus, $a = -1$ is indeed the unique real solution.

Q: Is there a shortcut to finding the distance from the origin?

While there are formulas for the distance of a point from the origin, they all rely on knowing the coordinates of that point. In this case, finding the intersection point is unavoidable. However, notice how the symmetry of the equations makes the final calculation of $\sqrt{x^2 + y^2}$ quite elegant. The beauty of the problem lies in the simplicity of the final result.

Question

If the two lines $x + (a - 1)y = 1$ and $2x + a^2y = 1 (a \in R - \{0, 1\})$ are perpendicular, then the distance of their point of intersection from the origin is :-

Options:

(A) $\frac{2}{5}$

(B) $\frac{2}{\sqrt{5}}$

(C) $\frac{\sqrt{2}}{5}$

(D) $\frac{\sqrt{2}}{\sqrt{5}}$

Sigma Percentile · Accepted Answer

### The Geometry of Perpendicularity: A Dance of Lines

Welcome, fellow explorer of the mathematical universe. Today, we are not just solving a problem; we are choreographing a dance between two lines on the Cartesian plane.

Imagine these lines as two dancers, their paths dictated by a mysterious parameter $a$. We are told they are perpendicular, meeting at a perfect $90^{\circ}$ angle.

Our goal is to find their meeting point and measure its distance from the origin. It is a journey of precision, logic, and algebraic elegance.

### Phase 1: The Slope Condition

Every line $Ax + By = C$ carries a secret: its slope. The slope $m$ is the ratio of the change in $y$ to the change in $x$, which for this standard form is $m = -\frac{A}{B}$.

For our first line $L_1: x + (a - 1)y = 1$, the slope is:
$$m_1 = -\frac{1}{a-1}$$

For our second line $L_2: 2x + a^2y = 1$, the slope is:
$$m_2 = -\frac{2}{a^2}$$

The condition for perpendicularity is a cornerstone of coordinate geometry: the product of the slopes of two perpendicular lines must be exactly $-1$. Mathematically, this is $m_1 \cdot m_2 = -1$.

### Phase 2: The Algebraic Hunt

Now, we substitute our slopes into the condition:
$$\left(-\frac{1}{a-1}\right) \cdot \left(-\frac{2}{a^2}\right) = -1$$

The negatives cancel out, leaving us with:
$$\frac{2}{a^2(a-1)} = -1$$

Cross-multiplying, we get $2 = -a^2(a-1)$, which expands to $2 = -a^3 + a^2$. Rearranging this, we arrive at the cubic equation:
$$a^3 - a^2 + 2 = 0$$

Do not let the cubic form intimidate you. In JEE problems, there is often a simple integer root waiting to be discovered. Testing $a = -1$, we find $(-1)^3 - (-1)^2 + 2 = -1 - 1 + 2 = 0$.

Factoring out $(a+1)$, we are left with $(a+1)(a^2 - 2a + 2) = 0$. The quadratic part has a negative discriminant, confirming that **$a = -1$** is our only real solution.

### Phase 3: The Intersection

With $a = -1$ locked in, our lines become concrete. Substituting $a = -1$ into $L_1$, we get $x - 2y = 1$.

For $L_2$, we get $2x + (-1)^2y = 1$, which simplifies to $2x + y = 1$. Now, we have a simple system of linear equations.

From $2x + y = 1$, we can express $y$ as $y = 1 - 2x$. Substituting this into $x - 2y = 1$:
$$x - 2(1 - 2x) = 1$$
$$x - 2 + 4x = 1$$
$$5x = 3$$

Thus, $x = \frac{3}{5}$. Substituting this back, $y = 1 - 2(\frac{3}{5}) = 1 - \frac{6}{5} = -\frac{1}{5}$. Our intersection point $P$ is $(\frac{3}{5}, -\frac{1}{5})$.

### Phase 4: The Final Distance

We have reached the final stage of our journey. We need the distance of point $P(\frac{3}{5}, -\frac{1}{5})$ from the origin $O(0,0)$.

The distance formula is $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$. Since one point is the origin, this simplifies to:
$$d = \sqrt{x^2 + y^2}$$

Plugging in our coordinates:
$$d = \sqrt{\left(\frac{3}{5}\right)^2 + \left(-\frac{1}{5}\right)^2} = \sqrt{\frac{9}{25} + \frac{1}{25}} = \sqrt{\frac{10}{25}} = \sqrt{\frac{2}{5}}$$

This is our final answer: **$\frac{\sqrt{10}}{5}$** (or $\sqrt{\frac{2}{5}}$).

### Conclusion

Look at what we have achieved. We started with two abstract lines and, through the power of the perpendicularity condition, we pinned down the parameter $a$, found the intersection point, and calculated the distance.

This is the essence of coordinate geometry: taking the abstract and making it concrete. Keep practicing this flow, and you will find that even the most complex problems become a series of logical, satisfying steps.

Animated Solution for Mathematics - Straight Lines: If the two lines $x + (a - 1) y = 1$ and $2 x + a^{2} y = 1 (a \in R - {0, 1})$ are perpendicular, then the distance of their point of intersection from the origin is :-

Select Answer:

Visualized Solution

Visualizing the Setup

Finding the Slopes

Applying Perpendicularity

Forming the Equation

Solving for $a$

Updating the Line Equations

Finding the Intersection Point $P$

Setting Up the Distance Formula

Calculating the Final Distance

Final Answer & Conclusion

The Sigma Insight: Angle Between Two Lines

The Geometry of Perpendicularity

Phase 1

Phase 2

Phase 3

Phase 4

Conclusion

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