Consider the lines x(3λ + 1) + y(7λ + 2) = 17λ + 5, λ being a parameter, all passing through a point

Q: Wait, why is the farthest line perpendicular to $OP$? Can't it be at an angle?

Great question! Imagine the origin $O$ and the fixed point $P$. Any line passing through $P$ forms a right-angled triangle with the origin, where the line itself is the hypotenuse of a triangle formed by the origin, the foot of the perpendicular, and $P$. The perpendicular distance from the origin to the line is one of the legs of this triangle. The hypotenuse $OP$ is always longer than any leg. The distance is maximized when the 'leg' becomes the hypotenuse itself, which happens only when the line is perpendicular to $OP$.

Q: Why can't we just differentiate the distance formula with respect to $\lambda$ to find the maximum?

You absolutely could! If you write the distance $d$ as a function of $\lambda$ and set $\frac{dd}{d\lambda} = 0$, you would find the same result. However, that path involves heavy algebraic manipulation and potential for calculation errors. The geometric approach—recognizing the perpendicularity condition—is the 'JEE way' because it saves time and demonstrates a deeper understanding of coordinate geometry.

Q: How do I know for sure that the line $L$ is $x + 2y - 5 = 0$?

We derived it using the point-slope form. We knew the line must pass through $P(1, 2)$ and have a slope of $-\frac{1}{2}$. Plugging these into $y - 2 = -\frac{1}{2}(x - 1)$ gives $2y - 4 = -x + 1$, which simplifies to $x + 2y - 5 = 0$. You can verify this by checking if $P(1, 2)$ satisfies the equation: $1 + 2(2) - 5 = 1 + 4 - 5 = 0$. It works perfectly!

Question

Consider the lines $x(3\lambda + 1) + y(7\lambda + 2) = 17\lambda + 5, \lambda$ being a parameter, all passing through a point $P$. One of these lines (say $L$) is farthest from the origin. If the distance of $L$ from the point $(3, 6)$ is $d$, then the value of $d^2$ is

Options:

(A) 20

(B) 30

(C) 10

(D) 15

Sigma Percentile · Accepted Answer

### The Dance of the Parameter: Unlocking the Family of Lines

Have you ever looked at an equation like $x(3\lambda + 1) + y(7\lambda + 2) = 17\lambda + 5$ and felt a shiver of intimidation? It is perfectly normal.

That $\lambda$ parameter feels like a moving target, a ghost in the machine that changes the line's orientation with every value it takes. But here is the secret: that $\lambda$ is not a source of chaos; it is a key.

In the world of coordinate geometry, this is a 'family of lines.' Imagine a pivot point, a nail driven into a board, and a ruler rotating around it. That nail is our fixed point $P$, and $\lambda$ is simply the force rotating the ruler.

Our first mission is to find where that nail is driven.

### The Fixed Point: The Anchor in the Storm

To find the fixed point $P$, we must strip away the complexity. We expand the equation:

$$3\lambda x + x + 7\lambda y + 2y = 17\lambda + 5$$

Now, we group the terms with $\lambda$ and the terms without. This gives us:

$$(x + 2y - 5) + \lambda(3x + 7y - 17) = 0$$

This is the classic $L_1 + \lambda L_2 = 0$ form. For this equation to hold true for *any* value of $\lambda$, both $L_1$ and $L_2$ must be zero simultaneously.

We are left with a simple system of linear equations:

$$x + 2y - 5 = 0$$
$$3x + 7y - 17 = 0$$

Solving these, we find $x = 1$ and $y = 2$. Our anchor point $P$ is $(1, 2)$.

### The Geometric Revelation: The Farthest Line

Now, let us step back and look at the big picture. We have a fixed point $P(1, 2)$ and the origin $O(0, 0)$. We are looking for the line $L$ passing through $P$ that is farthest from the origin.

Think about the geometry. If you draw a line segment $OP$, any line passing through $P$ will form a right-angled triangle with the origin, where the perpendicular distance from the origin to the line is one of the legs.

The segment $OP$ is the hypotenuse. Since the hypotenuse is always the longest side of a right triangle, the distance is maximized when the line itself is perpendicular to $OP$.

This is a beautiful, elegant realization that turns a complex optimization problem into a simple slope calculation.

### The Final Calculation: Bringing it Home

With the geometric insight in hand, the rest is a victory lap. The slope of $OP$ is:

$$m_{OP} = \frac{2 - 0}{1 - 0} = 2$$

Since our line $L$ is perpendicular to $OP$, its slope must be the negative reciprocal:

$$m_L = -\frac{1}{2}$$

Using the point-slope form $y - y_1 = m(x - x_1)$, we get $y - 2 = -\frac{1}{2}(x - 1)$, which simplifies to $x + 2y - 5 = 0$.

Finally, we calculate the distance $d$ from the point $Q(3, 6)$ to this line using the formula:

$$d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$$

Substituting our values, we get:

$$d = \frac{|(1)(3) + (2)(6) - 5|}{\sqrt{1^2 + 2^2}} = \frac{|3 + 12 - 5|}{\sqrt{5}} = \frac{10}{\sqrt{5}} = 2\sqrt{5}$$

The question asks for $d^2$, so:

$$d^2 = (2\sqrt{5})^2 = 4 	imes 5 = 20$$

We have conquered the problem, not by brute force, but by understanding the geometric soul of the equation. The final answer is **20**.

Select Answer:

Visualized Solution

The Family of Lines

Rearranging to $L_{1} + λ L_{2} = 0$

Finding the Fixed Point $P$

Solving for $P$

The Farthest Line Concept

Maximizing the Distance

Slopes of $O P$ and $L$

Equation of Line $L$

Distance from Point $Q (3, 6)$

Substituting Values for $d$

Calculating Distance $d$

Final Value of $d^{2}$

The Sigma Insight: Family of Lines

The Dance of the Parameter

The Fixed Point

The Geometric Revelation

The Final Calculation

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Select Answer:

Visualized Solution

The Family of Lines

Rearranging to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L1​+λL2​=0

Finding the Fixed Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

The Farthest Line Concept

Maximizing the Distance

Slopes of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } OP and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L

Equation of Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L

Distance from Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q(3,6)

Substituting Values for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d

Calculating Distance .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d

Final Value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d2