If the area of the triangle formed by the points `(2a,b), (a+b, 2b+a)` and (2b, 2a) be `Delta_(1)` and the area of the triangle whose vertices are `(a+b, a-b), (3b-a, b+3a)` and `(3a-b, 3b-a)` be `Delta_(2)`, then the value of `Delta_(2)//Delta_(1)` is

Answer

To find the value of \(\frac{\Delta_2}{\Delta_1}\), we will first calculate the areas \(\Delta_1\) and \(\Delta_2\) using the formula for the area of a triangle given its vertices.

### Step 1: Calculate \(\Delta_1\)

The vertices of the first triangle are given as:
- \(A(2a, b)\)
- \(B(a+b, 2b+a)\)
- \(C(2b, 2a)\)

Using the formula for the area of a triangle with vertices \((x_1, y_1)\), \((x_2, y_2)\), \((x_3, y_3)\):
\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]

Substituting the coordinates:
\[ \Delta_1 = \frac{1}{2} \left| 2a((2b + a) - 2a) + (a + b)(2a - b) + 2b(b - (2b + a)) \right| \]

Calculating each term:
1. \(2a((2b + a) - 2a) = 2a(2b - a) = 4ab - 2a^2\)
2. \((a + b)(2a - b) = 2a^2 + 2ab - ab - b^2 = 2a^2 + ab - b^2\)
3. \(2b(b - (2b + a)) = 2b(b - 2b - a) = 2b(-b - a) = -2b^2 - 2ab\)

Combining these:
\[ \Delta_1 = \frac{1}{2} \left| (4ab - 2a^2) + (2a^2 + ab - b^2) + (-2b^2 - 2ab) \right| \]
\[ = \frac{1}{2} \left| 4ab - 2ab - 2b^2 - b^2 \right| \]
\[ = \frac{1}{2} \left| 2ab - 3b^2 \right| \]

Thus, we have:
\[ \Delta_1 = \frac{1}{2} |2ab - 3b^2| \]

### Step 2: Calculate \(\Delta_2\)

The vertices of the second triangle are given as:
- \(D(a+b, a-b)\)
- \(E(3b-a, b+3a)\)
- \(F(3a-b, 3b-a)\)

Using the same area formula:
\[ \Delta_2 = \frac{1}{2} \left| (a+b)((b+3a) - (3b-a)) + (3b-a)((3b-a) - (a-b)) + (3a-b)((a-b) - (b+3a)) \right| \]

Calculating each term:
1. \((a+b)((b+3a) - (3b-a)) = (a+b)(4a - 2b)\)
2. \((3b-a)((3b-a) - (a-b)) = (3b-a)(4b - 2a)\)
3. \((3a-b)((a-b) - (b+3a)) = (3a-b)(-2b - 2a)\)

Calculating these terms separately:
1. \((a+b)(4a - 2b) = 4a^2 + 4ab - 2b^2 - 2ab = 4a^2 + 2ab - 2b^2\)
2. \((3b-a)(4b - 2a) = 12b^2 - 6ab - 4ab + 2a^2 = 12b^2 - 10ab + 2a^2\)
3. \((3a-b)(-2b - 2a) = -6ab - 6a^2 + 2b^2\)

Combining these:
\[ \Delta_2 = \frac{1}{2} \left| (4a^2 + 2ab - 2b^2) + (12b^2 - 10ab + 2a^2) + (-6ab - 6a^2 + 2b^2) \right| \]
\[ = \frac{1}{2} \left| 4a^2 + 2a^2 - 6a^2 + 2ab - 10ab - 6ab + 12b^2 - 2b^2 \right| \]
\[ = \frac{1}{2} \left| 0a^2 - 14ab + 10b^2 \right| \]

Thus, we have:
\[ \Delta_2 = \frac{1}{2} |10b^2 - 14ab| \]

### Step 3: Calculate \(\frac{\Delta_2}{\Delta_1}\)

Now we can find the ratio:
\[ \frac{\Delta_2}{\Delta_1} = \frac{\frac{1}{2} |10b^2 - 14ab|}{\frac{1}{2} |2ab - 3b^2|} = \frac{|10b^2 - 14ab|}{|2ab - 3b^2|} \]

### Final Result:
The value of \(\frac{\Delta_2}{\Delta_1}\) is:
\[ \frac{\Delta_2}{\Delta_1} = \frac{|10b^2 - 14ab|}{|2ab - 3b^2|} \]