Let the area of the triangle with vertices A(1, α), B(α, 0) and C(0, α) be 4 sq. units. If the point

Q: Why do we use the modulus in the area formula?

The area formula using coordinates gives the 'signed area'. Depending on the order of vertices (clockwise or counter-clockwise), the result can be negative. Since area must be positive, we take the absolute value.

Q: Can we use the determinant method instead?

Absolutely! The determinant method is just a compact way of writing the coordinate formula. It is often less prone to algebraic errors, but the result will be identical.

Q: Does the value of $\alpha$ (8 or -8) change the final answer for $\beta$?

Great question! Notice that in the final equation $\beta = -\alpha^2$, the term $\alpha^2$ is always positive regardless of whether $\alpha$ is $8$ or $-8$. Thus, $\beta$ remains $-64$.

Question

Let the area of the triangle with vertices $A(1, \alpha)$, $B(\alpha, 0)$ and $C(0, \alpha)$ be 4 sq. units. If the point $(\alpha, -\alpha)$, $(-\alpha, \alpha)$ and $(\alpha^2, \beta)$ are collinear, then $\beta$ is equal to

Options:

(A) 64

(B) -8

(C) -64

(D) 512

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on the coordinate plane, looking at a triangle defined by vertices $A(1, \alpha)$, $B(\alpha, 0)$, and $C(0, \alpha)$. We are given that the area of this triangle is $4$ square units.

To find the value of $\alpha$, we use the fundamental area formula for a triangle in coordinate geometry:

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

The modulus is a guardian of truth, ensuring the area remains positive regardless of vertex orientation. Substituting our coordinates into the formula, we obtain:

$$\frac{1}{2} |1(0 - \alpha) + \alpha(\alpha - \alpha) + 0(\alpha - 0)| = 4$$

The zeros simplify our work significantly, leaving us with the expression $\frac{1}{2} |-\alpha| = 4$. This simplifies to $|-\alpha| = 8$, which yields the two possible values:

$$\alpha = \pm 8$$

### The Elegance of Collinearity

Now, we shift our focus to the points $P(\alpha, -\alpha)$, $Q(-\alpha, \alpha)$, and $R(\alpha^2, \beta)$. We are given that these points are **collinear**, meaning they lie on a single, unbroken straight line.

Mathematically, this implies that the slope between any two pairs of points must be identical. We calculate the slope of $PQ$ as follows:

$$	ext{Slope}_{PQ} = \frac{\alpha - (-\alpha)}{-\alpha - \alpha} = \frac{2\alpha}{-2\alpha} = -1$$

This is a moment of mathematical elegance. The $\alpha$ terms cancel out, leaving us with a constant slope of $-1$, indicating the line is inclined at $135^\circ$ to the positive x-axis.

### The Final Synthesis

With the slope of $PQ$ established as $-1$, we set the slope of $QR$ equal to this value:

$$\frac{\beta - \alpha}{\alpha^2 - (-\alpha)} = -1$$

Multiplying both sides by the denominator $(\alpha^2 + \alpha)$, we get:

$$\beta - \alpha = -1 \cdot (\alpha^2 + \alpha)$$

$$\beta - \alpha = -\alpha^2 - \alpha$$

The $-\alpha$ terms on both sides cancel out perfectly, leaving us with the elegant result:

$$\beta = -\alpha^2$$

Since we previously determined that $\alpha = \pm 8$, it follows that $\alpha^2 = 64$. Substituting this value into our equation for $\beta$, we arrive at the final result:

$$\mathbf{\beta = -64}$$

Animated Solution for Mathematics - Straight Lines: Let the area of the triangle with vertices $A (1, α)$ , $B (α, 0)$ and $C (0, α)$ be 4 sq. units. If the point $(α, - α)$ , $(- α, α)$ and $(α^{2}, β)$ are collinear, then $β$ is equal to

Select Answer:

Visualized Solution

Visualizing Triangle $A B C$

The Area Formula

Substituting Coordinates

Simplifying the Expression

Solving for $α$

Collinearity of $P, Q, R$

Condition for Collinearity

Calculating Slope of $P Q$

Simplifying Slope $P Q$

Setting up Slope $QR$

Solving for $β$

Final Calculation

The Sigma Insight: Area of Triangle

Analyzing the Setup

The Elegance of Collinearity

The Final Synthesis

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Select Answer:

Visualized Solution

Visualizing Triangle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ABC

The Area Formula

Substituting Coordinates

Simplifying the Expression

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

Collinearity of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P,Q,R

Condition for Collinearity

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

Simplifying Slope .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PQ

Setting up Slope .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } QR

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