Let (α, β) be the centroid of the triangle formed by the lines 15x - y = 82, 6x - 5y = -4 and 9x + 4

Q: Why do we have to solve for all three vertices? Can't we just find two?

The centroid is the average of *all* three vertices. If you miss one, you miss the balance point of the entire triangle. Think of it like a see-saw; you need all weights to find the equilibrium. Every vertex contributes equally to the centroid's position.

Q: Is there a trick to avoid solving the systems of equations?

In competitive exams, sometimes there are shortcuts, but here, the lines are distinct and non-parallel. Solving the systems is the most robust path. Don't fear the algebra; embrace the precision. Each intersection is a unique coordinate pair waiting to be discovered.

Q: Why is the quadratic equation $x^2 - Sx + P = 0$?

It comes from the factored form $(x - r_1)(x - r_2) = 0$. Expanding this gives $x^2 - (r_1 + r_2)x + r_1 r_2 = 0$. It is the fundamental connection between roots and coefficients, a beautiful bridge between the values of the roots and the structure of the equation itself.

Question

Let $(\alpha, \beta)$ be the centroid of the triangle formed by the lines $15x - y = 82, 6x - 5y = -4$ and $9x + 4y = 17$. Then $\alpha + 2\beta$ and $2\alpha - \beta$ are the roots of the equation

Options:

(A) $x^2 - 7x + 12 = 0$

(B) $x^2 - 14x + 48 = 0$

(C) $x^2 - 13x + 42 = 0$

(D) $x^2 - 10x + 25 = 0$

Sigma Percentile · Accepted Answer

### The Geometry of Balance: Unveiling the Centroid

Welcome, future engineer. Today, we are not just solving a problem; we are embarking on a journey through the Cartesian plane.

We have three lines:
$L_1: 15x - y = 82$
$L_2: 6x - 5y = -4$
$L_3: 9x + 4y = 17$

These lines are the boundaries of a triangle. Our mission is to find its heart—the centroid—and then use that knowledge to construct a quadratic equation.

### Phase 1: The Hunt for Vertices

To find the centroid, we first need the vertices by solving the lines in pairs.

For vertex $A$, the intersection of $L_1$ and $L_2$, we use substitution. From $L_1$, we see $y = 15x - 82$. Substituting this into $L_2$ gives:

$$6x - 5(15x - 82) = -4$$

Expanding this, we get $6x - 75x + 410 = -4$, which simplifies to $-69x = -414$. Dividing both sides, we find $x = 6$. Plugging this back into our expression for $y$, we get $y = 15(6) - 82 = 8$. Thus, our first vertex is $A(6, 8)$.

Next, we find vertex $B$, the intersection of $L_2$ and $L_3$. We multiply $L_2$ by $4$ and $L_3$ by $5$ to align the $y$-coefficients:

$$24x - 20y = -16$$
$$45x + 20y = 85$$

Adding these, the $y$-terms vanish, leaving $69x = 69$, so $x = 1$. Substituting $x=1$ into $L_3$, we find $y = 2$. Vertex $B$ is $(1, 2)$.

Finally, for vertex $C$, the intersection of $L_1$ and $L_3$, we substitute $y = 15x - 82$ into $L_3$:

$$9x + 4(15x - 82) = 17$$

This expands to $9x + 60x - 328 = 17$, or $69x = 345$. Dividing by $69$, we get $x = 5$. Then, $y = 15(5) - 82 = -7$. Vertex $C$ is $(5, -7)$.

### Phase 2: The Centroid

Now that we have our vertices $A(6, 8)$, $B(1, 2)$, and $C(5, -7)$, we seek the centroid $(\alpha, \beta)$. The centroid is the average of the vertices:

$$\alpha = \frac{x_1 + x_2 + x_3}{3}, \quad \beta = \frac{y_1 + y_2 + y_3}{3}$$

Calculating $\alpha$:

$$\alpha = \frac{6 + 1 + 5}{3} = \frac{12}{3} = 4$$

Calculating $\beta$:

$$\beta = \frac{8 + 2 + (-7)}{3} = \frac{3}{3} = 1$$

Our centroid is $(4, 1)$.

### Phase 3: The Quadratic Bridge

The problem defines two roots, $r_1 = \alpha + 2\beta$ and $r_2 = 2\alpha - \beta$. Substituting $\alpha = 4$ and $\beta = 1$:

$$r_1 = 4 + 2(1) = 6$$
$$r_2 = 2(4) - 1 = 7$$

We now have the roots of our quadratic equation: $6$ and $7$. To form the equation, we calculate the sum of roots $S$ and the product of roots $P$:

$$S = 6 + 7 = 13$$
$$P = 6 	imes 7 = 42$$

The standard form is $x^2 - Sx + P = 0$. Therefore, the final equation is:

**$x^2 - 13x + 42 = 0$**

Animated Solution for Mathematics - Straight Lines: Let $(α, β)$ be the centroid of the triangle formed by the lines $15 x - y = 82, 6 x - 5 y = - 4$ and $9 x + 4 y = 17$ . Then $α + 2 β$ and $2 α - β$ are the roots of the equation

Select Answer:

Visualized Solution

Visualizing the Triangle

Finding Vertex $A$

Finding Vertex $B$

Finding Vertex $C$

The Centroid Formula

Calculating the Centroid $(α, β)$

Defining the Roots

Calculating the Roots

Forming the Quadratic Equation

The Sigma Insight: Centroid, Incenter, Orthocenter, and Circumcenter

The Geometry of Balance

Phase 1

Phase 2

Phase 3

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