In an isosceles triangle ABC, the vertex A is (6, 1) and the equation of the base BC is 2x + y = 4.

Q: Wait, why do we reject the root $h = 1$ in the quadratic equation?

Great catch! When we solve the quadratic equation $5h^2 - 24h + 19 = 0$, we get two roots: $h = 1$ and $h = \frac{19}{5}$. If we plug $h = 1$ into our parametric form for $C$, we get $(1, 4 - 2(1)) = (1, 2)$, which is exactly the coordinate of point $B$. Since $C$ must be a distinct vertex of the triangle, we must discard this solution.

Q: Why can't we just use the property that the altitude from $A$ bisects $BC$?

You absolutely can! That is a valid geometric approach. You would find the slope of $BC$, then the slope of the altitude from $A$, find the equation of the altitude, find the intersection point $M$ on $BC$, and then use the midpoint formula to find $C$. However, the distance formula approach is often faster and less prone to calculation errors in coordinate geometry problems like this.

Q: Why did we use $AB^2$ instead of $AB$?

Using $AB^2$ is a standard trick in coordinate geometry. It allows us to work with the distance formula without dealing with square roots, which makes the algebra much cleaner and prevents potential errors when squaring both sides later.

Question

In an isosceles triangle $ABC$, the vertex $A$ is $(6, 1)$ and the equation of the base $BC$ is $2x + y = 4$. Let the point $B$ lie on the line $x + 3y = 7$. If $(\alpha, \beta)$ is the centroid $\Delta ABC$, then $15(\alpha + \beta)$ is equal to :

Options:

(A) 39

(B) 41

(C) 51

(D) 63

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a coordinate plane. You have a fixed point $A$ at $(6, 1)$, and a line $2x + y = 4$ stretching out like a horizon, representing the base of a triangle.

We are tasked with finding the centroid of an isosceles triangle $ABC$, where the base $BC$ lies on that line, and vertex $B$ is constrained to another line, $x + 3y = 7$.

### Locating the Anchor

Point $B$ is the intersection of our base line $2x + y = 4$ and the constraint line $x + 3y = 7$. To find this, we solve the system of equations.

From the first, we have $y = 4 - 2x$. Substituting this into the second, we get:

$$x + 3(4 - 2x) = 7$$

This simplifies to $x + 12 - 6x = 7$, or $-5x = -5$. Thus, $x = 1$ and $y = 2$. Our anchor point $B$ is at $(1, 2)$.

### The Isosceles Constraint

We know $A(6, 1)$ and $B(1, 2)$. The distance $AB$ is the length of one of the equal legs of our isosceles triangle. Using the distance formula:

$$AB^2 = (6 - 1)^2 + (1 - 2)^2 = 5^2 + (-1)^2 = 26$$

Because the triangle is isosceles with base $BC$, we know that $AC = AB$, which means $AC^2 = 26$ as well. This is our golden key.

### The Power of Parametrization

We need to find point $C$. Since $C$ lies on the line $2x + y = 4$, we can define its coordinates using a single parameter $h$.

Let $x_C = h$. Then $y_C = 4 - 2h$. So, $C = (h, 4 - 2h)$.

Now, we apply the distance formula again for $AC^2$:

$$(h - 6)^2 + (4 - 2h - 1)^2 = 26$$

Simplifying the second term, we get:

$$(h - 6)^2 + (3 - 2h)^2 = 26$$

### Solving the Quadratic

Expanding this, we get:

$$(h^2 - 12h + 36) + (9 - 12h + 4h^2) = 26$$

Combining like terms, we arrive at:

$$5h^2 - 24h + 45 = 26 \Rightarrow 5h^2 - 24h + 19 = 0$$

Factoring this quadratic, we find $(h - 1)(5h - 19) = 0$. This gives us two potential values for $h$: $1$ and $\frac{19}{5}$.

Since $h=1$ corresponds to point $B$, we must choose $h = \frac{19}{5}$. Substituting this back, we find:

$$C = \left( \frac{19}{5}, 4 - 2\left( \frac{19}{5} \right) \right) = \left( \frac{19}{5}, -\frac{18}{5} \right)$$

### Final Calculation

The centroid $(\alpha, \beta)$ is the average of the vertices:

$$\alpha = \frac{6 + 1 + \frac{19}{5}}{3} = \frac{7 + \frac{19}{5}}{3} = \frac{54}{15}$$

$$\beta = \frac{1 + 2 - \frac{18}{5}}{3} = \frac{3 - \frac{18}{5}}{3} = -\frac{3}{15}$$

The question asks for $15(\alpha + \beta)$. Substituting our values:

$$15 \left( \frac{54}{15} - \frac{3}{15} \right) = 15 \left( \frac{51}{15} \right) = 51$$

We have navigated the geometry, tamed the algebra, and arrived at the final answer: **51**.

Animated Solution for Mathematics - Straight Lines: In an isosceles triangle $A B C$ , the vertex $A$ is $(6, 1)$ and the equation of the base $B C$ is $2 x + y = 4$ . Let the point $B$ lie on the line $x + 3 y = 7$ . If $(α, β)$ is the centroid $Δ A B C$ , then $15 (α + β)$ is equal to :

Select Answer:

Visualized Solution

Visualizing the Geometry

Finding Point $B$

Calculating Distance $A B^{2}$

Parametric Form of Point $C$

Setting up $A C^{2} = A B^{2}$

Expanding the Equation

Solving for $h$

Coordinates of $C$

Finding the Centroid $(α, β)$

Final Calculation

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

Analyzing the Setup

Locating the Anchor

The Isosceles Constraint

The Power of Parametrization

Solving the Quadratic

Final Calculation

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Visualized Solution

Visualizing the Geometry

Finding Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B

Calculating Distance .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AB2

Parametric Form of Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

Setting up .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AC2=AB2

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Finding the Centroid .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (α,β)

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

Analyzing the Setup

Locating the Anchor

The Isosceles Constraint

The Power of Parametrization

Solving the Quadratic

Final Calculation

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