Let A be the set of all points (α, β) such that the area of triangle formed by the points (5, 6), (3

Q: Wait, why do we get two lines instead of just one?

Great observation! The area formula involves an absolute value: $|2x - y - 4| = 12$. Geometrically, this means the third vertex can be on either side of the line segment connecting $(5, 6)$ and $(3, 2)$ while maintaining the same perpendicular height. Thus, we get two parallel lines, one on each side.

Q: Can I just pick a random point on the line to find the distance?

You could, but that would be incredibly tedious! The shortest distance from a point to a line is always the perpendicular distance. Using the formula $d = \frac{|c|}{\sqrt{a^2 + b^2}}$ is the most elegant and efficient way to ensure you find the absolute minimum.

Q: Why is the line parallel to the segment connecting the two fixed points?

Think about the area formula: $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$. Since the base (the distance between $(5, 6)$ and $(3, 2)$) is fixed, the height must also be fixed for the area to remain 12. A constant height relative to a line defines a parallel line.

Q: Is it possible for the origin to be between the two lines?

Yes, it is! If you look at the equations $2x - y - 16 = 0$ and $2x - y + 8 = 0$, the origin $(0,0)$ yields $-16$ for the first and $+8$ for the second. Since the signs are different, the origin lies between the two lines. This is why we calculate both distances and choose the smaller one.

Q: What if I made a calculation error in the area formula expansion?

Algebraic slips are the most common trap in JEE Advanced. Always double-check your signs inside the determinant or the coordinate formula. If you get a weird result, re-verify the expansion of $5(2 - \beta) + 3(\beta - 6) + \alpha(6 - 2)$ carefully.

Question

Let $A$ be the set of all points $(\alpha, \beta)$ such that the area of triangle formed by the points $(5, 6), (3, 2)$ and $(\alpha, \beta)$ is 12 square units. Then the least possible length of a line segment joining the origin to a point in $A$, is :

Options:

(A) $\frac{4}{\sqrt{5}}$

(B) $\frac{16}{\sqrt{5}}$

(C) $\frac{8}{\sqrt{5}}$

(D) $\frac{12}{\sqrt{5}}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a vast coordinate plane. You have two fixed anchors: point $P_1(5, 6)$ and point $P_2(3, 2)$.

You are tasked with placing a third point $P_3(\alpha, \beta)$ such that the triangle formed by these three points always has an area of exactly $12$ square units. As you move $P_3$ around, you realize it is tracing a very specific path, uncovering the hidden geometry behind an algebraic constraint.

### The Algebraic Foundation

To find this path, we turn to the classic area formula for a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$:

$$	ext{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$$

Substituting our known points $(5, 6)$ and $(3, 2)$ and our variable point $(\alpha, \beta)$, we get:

$$\frac{1}{2} |5(2 - \beta) + 3(\beta - 6) + \alpha(6 - 2)| = 12$$

Expanding the terms inside the absolute value gives us $|10 - 5\beta + 3\beta - 18 + 4\alpha| = 24$. Combining like terms, we arrive at $|4\alpha - 2\beta - 8| = 24$.

Dividing by $2$, we find the soul of our locus: $|2\alpha - \beta - 4| = 12$. This represents two parallel lines, $2x - y - 16 = 0$ and $2x - y + 8 = 0$, where we have replaced $(\alpha, \beta)$ with $(x, y)$ for clarity.

### The Quest for the Minimum

Now, the problem shifts. We need to find the least possible distance from the origin $(0,0)$ to any point on these lines.

We use the distance formula for a point $(x_0, y_0)$ to a line $ax + by + c = 0$:

$$d = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}$$

For our first line, $2x - y - 16 = 0$, the distance is:

$$d_1 = \frac{|-16|}{\sqrt{2^2 + (-1)^2}} = \frac{16}{\sqrt{5}}$$

For our second line, $2x - y + 8 = 0$, the distance is:

$$d_2 = \frac{|8|}{\sqrt{2^2 + (-1)^2}} = \frac{8}{\sqrt{5}}$$

### The Final Revelation

We are looking for the **least** possible length. Comparing $d_1 = \frac{16}{\sqrt{5}}$ and $d_2 = \frac{8}{\sqrt{5}}$, it is clear that the second line is closer to the origin.

The area constraint effectively traps our point on two parallel tracks, and the origin simply picks the track that is closer to it. The final answer is:

$$\mathbf{\frac{8}{\sqrt{5}}}$$

Animated Solution for Mathematics - Straight Lines: Let $A$ be the set of all points $(α, β)$ such that the area of triangle formed by the points $(5, 6), (3, 2)$ and $(α, β)$ is 12 square units. Then the least possible length of a line segment joining the origin to a point in $A$ , is :

Select Answer:

Visualized Solution

Visualizing the Fixed Points

The Variable Point and Triangle Area

Applying the Area Formula

Simplifying the Expression

Refining the Locus Equation

Splitting into Two Cases

Visualizing the Locus Lines

The Objective: Shortest Distance

Distance Formula Setup

Calculating Distance to $L_{1}$

Calculating Distance to $L_{2}$

Comparing and Concluding

The Sigma Insight: Distance of a Point from a Line

Analyzing the Setup

The Algebraic Foundation

The Quest for the Minimum

The Final Revelation

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