Plz answer question 12

The centroid of an equilateral triangle is (0,0) if two vertices of the triangle lie on x+y=2root 2 then one of them will have coordinates

To find the coordinates of one vertex of an equilateral triangle given that its centroid is at (0,0) and two of its vertices lie on the line represented by the equation x + y = 2√2, we can approach this problem step by step.

Understanding the Centroid of a Triangle

The centroid of a triangle is the point where all three medians intersect, and it is also the average of the vertices' coordinates. For a triangle with vertices A(x1, y1), B(x2, y2), and C(x3, y3), the coordinates of the centroid (G) can be calculated as:

• Gx = (x1 + x2 + x3) / 3
• Gy = (y1 + y2 + y3) / 3

In our case, since the centroid is at (0,0), we have:

• 0 = (x1 + x2 + x3) / 3
• 0 = (y1 + y2 + y3) / 3

Setting Up the Problem

Let’s denote the two vertices on the line x + y = 2√2 as A(x1, y1) and B(x2, y2). The third vertex C(x3, y3) will be determined based on the properties of the centroid. Since the centroid is at (0,0), we can express the coordinates of C in terms of A and B:

• x3 = - (x1 + x2)
• y3 = - (y1 + y2)

Finding the Vertices

Since points A and B lie on the line x + y = 2√2, we can express y1 and y2 in terms of x1 and x2:

• y1 = 2√2 - x1
• y2 = 2√2 - x2

Substituting these values into the equations for the centroid gives us:

• 0 = (x1 + x2 - (2√2 - x1) - (2√2 - x2)) / 3

Rearranging this, we find:

• 0 = (x1 + x2 - 2√2 + x1 - 2√2 + x2) / 3
• 0 = (2x1 + 2x2 - 4√2) / 3

Simplifying the Equation

Multiplying through by 3 gives:

• 2x1 + 2x2 - 4√2 = 0

This simplifies to:

• x1 + x2 = 2√2

Finding One Vertex

Now, we can choose a specific value for x1 to find the corresponding y1 and then use x2 = 2√2 - x1 to find y2. Let’s choose x1 = √2:

• y1 = 2√2 - √2 = √2

This gives us the coordinates for vertex A:

• A(√2, √2)

Now, using x2 = 2√2 - √2 = √2, we have:

• B(√2, √2)

However, since we need distinct vertices, let’s choose another value for x1, say x1 = 2√2 - 1:

• y1 = 2√2 - (2√2 - 1) = 1

This gives us:

• A(2√2 - 1, 1)

Now, we can find B using x2 = 1:

• B(1, 2√2 - 1)

Final Vertex Calculation

The third vertex C can now be calculated as:

• x3 = -((2√2 - 1) + 1) = -2√2 + 1
• y3 = -((1) + (2√2 - 1)) = -2√2

Conclusion

Thus, we have one of the vertices as A(2√2 - 1, 1) or you could choose A(1, 2√2 - 1) depending on your choice of x1. Both satisfy the conditions laid out by the problem.