Let the points (11/2, α) lie on or inside the triangle with sides x+y=11, x+2y=16 and 2x+3y=29 Then

Q: Wait, why do we need to find the vertices first? Can't we just plug in $x = 5.5$ into all three lines?

That is a great question! While plugging in $x = 5.5$ into all three lines gives you three potential $y$-values, you need to know which two lines actually bound the triangle at that specific $x$-coordinate. Finding the vertices allows you to visualize the triangle and correctly identify the upper and lower boundaries for the vertical slice at $x = 5.5$.

Q: How do I know which lines are the upper and lower boundaries?

Once you have the vertices, you can see the triangle's shape. For any vertical line $x = k$ that passes through the triangle, the $y$-values inside the triangle are bounded by the two lines that form the sides of the triangle in that $x$-interval. By checking the $x$-coordinates of the vertices, you can determine which sides are relevant for your specific $x$-value.

Q: Why did we reject one of the lines when calculating the range of $\alpha$?

Excellent observation! The vertical line $x = 5.5$ only intersects two of the three sides of the triangle. The third side exists in a different $x$-range. Since our point is constrained to $x = 5.5$, we only care about the boundaries that actually exist at that $x$-coordinate. The third line is irrelevant for this specific slice of the triangle.

Question

Let the points $(\frac{11}{2},\alpha)$ lie on or inside the triangle with sides $x+y=11$, $x+2y=16$ and $2x+3y=29$ Then the product of the smallest and the largest values of a is equal to:

Options:

(A) 44

(B) 22

(C) 33

(D) 55

Sigma Percentile · Accepted Answer

### The Geometry of Constraints

Imagine you are standing on a coordinate plane, looking at three lines that slice across the grid. These lines are not just arbitrary equations; they are the walls of a triangular room.

Our goal is to find the range of possible heights, or $y$-values, for a point that is trapped inside this room at a specific horizontal position. This is the essence of coordinate geometry—turning abstract algebraic constraints into a tangible, visual reality.

### Phase 1: Mapping the Territory

Before we can understand the point $(\frac{11}{2}, \alpha)$, we must first understand the room itself. The triangle is defined by three lines:

$$L_1: x + y = 11$$
$$L_2: x + 2y = 16$$
$$L_3: 2x + 3y = 29$$

To find the corners of our room, we need to find where these walls meet. This is a classic exercise in solving systems of linear equations.

By taking the intersection of $L_1$ and $L_2$, we find Vertex $A = (6, 5)$. By intersecting $L_1$ and $L_3$, we find Vertex $B = (4, 7)$.

Finally, by intersecting $L_2$ and $L_3$, we find Vertex $C = (10, 3)$. Now, we have a clear map of our triangular region with vertices at $(6, 5)$, $(4, 7)$, and $(10, 3)$.

### Phase 2: The Vertical Slice

The problem places our point at $x = \frac{11}{2}$, which is $5.5$. This means our point is restricted to a vertical line slicing through the triangle.

Looking at our vertices, the $x$-coordinates range from $4$ to $10$. Since $5.5$ falls between $4$ and $6$, our vertical line passes through the triangle in the region between Vertex $B$ and Vertex $A$.

This is the crucial insight: the vertical line $x = 5.5$ intersects the sides $AB$ and $BC$, but it does not touch side $AC$ because that side exists in the $x$-range of $6$ to $10$.

### Phase 3: Trapping the Variable

Now that we know our vertical line intersects sides $AB$ and $BC$, we can find the boundaries for $\alpha$. Side $AB$ is part of line $L_1$, and side $BC$ is part of line $L_3$.

To find the smallest value of $\alpha$, we look at the lower boundary, which is line $L_1$ ($x + y = 11$). Substituting $x = 5.5$ into $x + y = 11$, we get $5.5 + y = 11$, which simplifies to $y = 5.5$.

This is our smallest $\alpha$.

To find the largest value of $\alpha$, we look at the upper boundary, which is line $L_3$ ($2x + 3y = 29$). Substituting $x = 5.5$ into $2x + 3y = 29$, we get:

$$2(5.5) + 3y = 29$$
$$11 + 3y = 29$$
$$3y = 18$$
$$y = 6$$

This is our largest $\alpha$.

### The Final Elegance

We have successfully trapped $\alpha$ between $5.5$ and $6$. The problem asks for the product of these two values.

Calculating the product, we have:

$$\frac{11}{2} 	imes 6 = 11 	imes 3 = 33$$

It is a beautiful result, demonstrating how a seemingly complex problem about a point inside a triangle can be broken down into simple, logical steps of intersection and boundary analysis. The final answer is **33**.

Animated Solution for Mathematics - Straight Lines: Let the points $(\frac{11}{2}, α)$ lie on or inside the triangle with sides $x + y = 11$ , $x + 2 y = 16$ and $2 x + 3 y = 29$ Then the product of the smallest and the largest values of a is equal to:

Select Answer:

Visualized Solution

The Boundary Lines

Finding Vertex A

Coordinates of Vertex A

Finding Vertex B

Coordinates of Vertex B

Finding Vertex C

Coordinates of Vertex C

Locating the Point

Boundaries at $x = 5.5$

Smallest Value of $α$

Largest Value of $α$

Final Product

The Sigma Insight: Various Forms of Equations of a Line

The Geometry of Constraints

Phase 1

Phase 2

Phase 3

The Final Elegance

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