Let B and C be the two points on the line y + x = 0 such that B and C are symmetric with respect to

Q: Wait, why do we assume the origin is the midpoint of $BC$? The problem just says they are symmetric with respect to the origin.

That is a great question! In coordinate geometry, when two points are symmetric with respect to a point (in this case, the origin), that point acts as the midpoint of the segment connecting them. Since the origin is $(0,0)$, it is the midpoint of $BC$ by definition of symmetry.

Q: Why is the area formula $\frac{h^2}{\sqrt{3}}$? I only know the formula $\frac{\sqrt{3}}{4}a^2$ where $a$ is the side length.

Both formulas are correct! The standard formula is $Area = \frac{\sqrt{3}}{4}a^2$. In an equilateral triangle, the altitude $h$ is related to the side $a$ by $h = \frac{\sqrt{3}}{2}a$. If you solve for $a$ ($a = \frac{2h}{\sqrt{3}}$) and substitute it into the standard area formula, you get $Area = \frac{\sqrt{3}}{4} (\frac{2h}{\sqrt{3}})^2 = \frac{\sqrt{3}}{4} \cdot \frac{4h^2}{3} = \frac{h^2}{\sqrt{3}}$. It is a very handy shortcut for problems where you already know the altitude!

Q: How do I know for sure that vertex $A$ lies on the altitude line $y = x$?

Think about the geometry of an equilateral triangle. The altitude from a vertex to the opposite side is also the median. Since the origin $O$ is the midpoint of $BC$, the line segment $AO$ is the median. Therefore, $AO$ must be the altitude. Since $A$ is a vertex and $O$ is the midpoint of the opposite side, $A$ must lie on the line containing the altitude $AO$.

Question

Let B and C be the two points on the line $y + x = 0$ such that B and C are symmetric with respect to the origin. Suppose A is a point on $y - 2x = 2$ such that $\Delta ABC$ is an equilateral triangle. Then, the area of the $\Delta ABC$ is

Options:

(A) $3\sqrt{3}$

(B) $2\sqrt{3}$

(C) $\frac{8}{\sqrt{3}}$

(D) $\frac{10}{\sqrt{3}}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow explorer of the mathematical landscape. Today, we are going to dissect a problem that, at first glance, might seem like a standard coordinate geometry exercise, but is actually a beautiful dance of symmetry and perpendicularity.

We are tasked with finding the area of an equilateral triangle $\Delta ABC$, where the base $BC$ rests on the line $x + y = 0$, and the third vertex $A$ is constrained to the line $y - 2x = 2$.

### The Mirror of the Origin

Imagine you are standing on the Cartesian plane. You see the line $L_1: x + y = 0$ passing through the origin.

The problem states that points $B$ and $C$ lie on this line and are symmetric with respect to the origin. In the language of geometry, symmetry about the origin means the origin $O(0,0)$ is the midpoint of the segment $BC$. This point is the anchor of our entire triangle.

### The Altitude's Path

Recall the properties of an equilateral triangle. In any equilateral triangle, the median drawn from a vertex to the opposite side is also the altitude.

Since $O$ is the midpoint of $BC$, the line segment $AO$ is the median. Consequently, $AO$ must be the altitude, meaning $AO \perp BC$.

The slope of our base line $BC$ ($x + y = 0$) is $m_1 = -1$. Since $AO$ is perpendicular to $BC$, its slope $m_2$ must be the negative reciprocal of $m_1$. Thus, $m_2 = 1$.

A line passing through the origin $(0,0)$ with a slope of $1$ has the beautifully simple equation:

$$y = x$$

### The Intersection

We now have two lines that define the position of vertex $A$. We know $A$ lies on the line $y - 2x = 2$, and we have just discovered that $A$ must also lie on the altitude line $y = x$.

Vertex $A$ is trapped at the intersection of these two paths. To find its coordinates, we substitute $y = x$ into the equation of the second line:

$$x - 2x = 2 \implies -x = 2 \implies x = -2$$

Since $y = x$, it follows that $y = -2$. Thus, our vertex $A$ is located at the point $(-2, -2)$.

### Final Calculation

With $A(-2, -2)$ and the origin $O(0,0)$ identified, the altitude $h$ of our triangle is the distance between these two points. Using the distance formula:

$$h = \sqrt{(-2 - 0)^2 + (-2 - 0)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}$$

We can use the altitude-based formula for an equilateral triangle, which is $Area = \frac{h^2}{\sqrt{3}}$. Substituting our value for $h$:

$$Area = \frac{(2\sqrt{2})^2}{\sqrt{3}} = \frac{8}{\sqrt{3}}$$

The area of $\Delta ABC$ is **$\frac{8}{\sqrt{3}}$** square units. It is a testament to how, by simply understanding the geometric properties of symmetry and perpendicularity, we can navigate through what initially seemed like a complex problem.

Animated Solution for Mathematics - Straight Lines: Let B and C be the two points on the line $y + x = 0$ such that B and C are symmetric with respect to the origin. Suppose A is a point on $y - 2 x = 2$ such that $Δ A B C$ is an equilateral triangle. Then, the area of the $Δ A B C$ is

Select Answer:

Visualized Solution

The Base Line $x + y = 0$

Symmetry about the Origin

Altitude of Equilateral $Δ A B C$

Equation of Altitude $A O$

Locating Vertex $A$

Solving for Coordinates of $A$

Visualizing the Complete Triangle

Calculating Altitude Length $h$

Area Formula for Equilateral Triangle

Final Area Calculation

The Sigma Insight: Various Forms of Equations of a Line

Analyzing the Setup

The Mirror of the Origin

The Altitude's Path

The Intersection

Final Calculation

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