Determine all values of alpha for which the point (alpha, alpha^2) lies inside the triangle formed b

Q: Why do we check the opposite vertex? Can't we just pick any point?

That is a fantastic question! The opposite vertex is your 'anchor'. It is guaranteed to be on the correct side of the line because it is a corner of the triangle. If your point $P$ is on the same side as that vertex, it is on the 'inside' of the triangle relative to that line. If you picked a random point, you wouldn't know if it was inside or outside, so the vertex is the only reliable reference point.

Q: What if the inequality results in a range that includes the boundary?

Remember, the problem asks for the point to be *inside* the triangle. This implies strict inequalities. If the boundary is included, the point is on the edge, not inside. Always use strict inequalities ($>$ or $<$) unless the problem specifies 'inside or on the boundary'. In this case, we exclude the roots of the quadratic equations.

Q: How do I handle the intersection of three different intervals? It gets confusing.

I know it feels like a mess, but here is the trick: use a number line. Draw one line for each inequality. Mark the regions that satisfy the condition. Then, look for the vertical strip where all three lines have a shaded region. That overlap is your final answer. Don't try to do it in your head; draw it out!

Question

Determine all values of $\alpha$ for which the point $(\alpha, \alpha^2)$ lies inside the triangle formed by the lines $2x + 3y - 1 = 0$, $x + 2y - 3 = 0$, $5x - 6y - 1 = 0$.

Sigma Percentile · Accepted Answer

### The Dance of the Parabola and the Triangle: A Geometric Journey Welcome, future engineer. Today, we are not just solving a coordinate geometry problem; we are witnessing a beautiful, precise dance between a curve and a polygon. We have a moving point $P$ defined by the coordinates $(\alpha, \alpha^2)$. This point is shackled to the path of the standard parabola $y = x^2$. Our mission is to find the specific values of $\alpha$ that allow this point to reside strictly within the sanctuary of a triangle formed by three lines: $L_1: 2x + 3y - 1 = 0$, $L_2: x + 2y - 3 = 0$, and $L_3: 5x - 6y - 1 = 0$. ### Mapping the Territory Before we dive into the algebra, we must understand our boundaries. A triangle is defined by its vertices, which we find by solving the pairwise intersections of our lines. Solving $L_1$ and $L_2$ gives us vertex $A(-7, 5)$. Solving $L_2$ and $L_3$ gives us vertex $B(\frac{5}{4}, \frac{7}{8})$. Solving $L_1$ and $L_3$ gives us vertex $C(\frac{1}{3}, \frac{1}{9})$. These three points are the pillars of our triangle. They define the cage within which our point $P$ must live. ### The 'Same-Side' Condition The core of the problem lies in defining 'inside' mathematically. If you are inside a triangle, you must be on the same side of each boundary line as the vertex that is opposite to that line. If we define our lines as functions $L(x, y) = ax + by + c$, then for a point $P$ to be inside, the sign of $L(P)$ must match the sign of $L( ext{opposite vertex})$. ### The Algebraic Siege Let us tackle this line by line. For $L_1: 2x + 3y - 1 = 0$, the opposite vertex is $B(\frac{5}{4}, \frac{7}{8})$. Plugging $B$ into $L_1$, we get: $$2\left(\frac{5}{4} ight) + 3\left(\frac{7}{8} ight) - 1 = \frac{33}{8} > 0$$ Thus, for $P(\alpha, \alpha^2)$ to be inside, we need $2\alpha + 3\alpha^2 - 1 > 0$. Factoring this gives $(3\alpha - 1)(\alpha + 1) > 0$, which yields the interval: $$\alpha \in (-\infty, -1) \cup \left(\frac{1}{3}, \infty ight)$$ Next, for $L_2: x + 2y - 3 = 0$, the opposite vertex is $C(\frac{1}{3}, \frac{1}{9})$. Plugging $C$ into $L_2$, we get: $$\frac{1}{3} + 2\left(\frac{1}{9} ight) - 3 = -\frac{22}{9} < 0$$ Thus, we need $x + 2y - 3 < 0$. Substituting $P$, we get $\alpha + 2\alpha^2 - 3 < 0$, or $(2\alpha + 3)(\alpha - 1) < 0$. This gives the interval: $$\alpha \in \left(-\frac{3}{2}, 1 ight)$$ Finally, for $L_3: 5x - 6y - 1 = 0$, the opposite vertex is $A(-7, 5)$. Plugging $A$ into $L_3$, we get: $$5(-7) - 6(5) - 1 = -66 < 0$$ Thus, we need $5x - 6y - 1 < 0$. Substituting $P$, we get $5\alpha - 6\alpha^2 - 1 < 0$, which rearranges to $6\alpha^2 - 5\alpha + 1 > 0$. Factoring this gives $(3\alpha - 1)(2\alpha - 1) > 0$, yielding the interval: $$\alpha \in \left(-\infty, \frac{1}{3} ight) \cup \left(\frac{1}{2}, \infty ight)$$ ### The Final Intersection We must find the intersection of our three conditions: 1. $\alpha \in (-\infty, -1) \cup (\frac{1}{3}, \infty)$ 2. $\alpha \in (-\frac{3}{2}, 1)$ 3. $\alpha \in (-\infty, \frac{1}{3}) \cup (\frac{1}{2}, \infty)$ The intersection of the first two conditions gives $(-\frac{3}{2}, -1) \cup (\frac{1}{3}, 1)$. When we intersect this result with the third condition, the region $(\frac{1}{3}, \frac{1}{2})$ is excluded. We are left with the final, elegant solution: $$\alpha \in \left(-\frac{3}{2}, -1 ight) \cup \left(\frac{1}{2}, 1 ight)$$ Take a moment to appreciate this. We have successfully constrained a moving point to a specific, finite region of space using nothing but algebraic inequalities. This is the power of coordinate geometry—turning visual intuition into rigorous, undeniable truth.

Animated Solution for Mathematics - Straight Lines: Determine all values of $α$ for which the point $(α, α^{2})$ lies inside the triangle formed by the lines $2 x + 3 y - 1 = 0$ , $x + 2 y - 3 = 0$ , $5 x - 6 y - 1 = 0$ .

Visualized Solution

Visualizing the Setup

Finding the Vertices of the Triangle

The 'Same Side' Condition

Condition for Line $L_{1}$

Solving the Inequality for $L_{1}$

Condition for Line $L_{2}$

Solving the Inequality for $L_{2}$

Condition for Line $L_{3}$

Solving the Inequality for $L_{3}$

Finding the Common Intersection

The Sigma Insight: Various Forms of Equations of a Line

The Dance of the Parabola and the Triangle

Mapping the Territory

The 'Same-Side' Condition

The Algebraic Siege

The Final Intersection

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