If the point P(3, 4) is equidistant from the points A`(a+b, b-a) and B (a-b, a+b) ` then prove that `3b-4a=0` .

Answer

To prove that the point P(3, 4) is equidistant from the points A(a+b, b-a) and B(a-b, a+b), we will use the distance formula and equate the distances PA and PB.

### Step-by-Step Solution:

1. **Distance Formula**: The distance \(d\) between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

2. **Calculate PA**: The distance from point P(3, 4) to point A(a+b, b-a) is:
\[ PA = \sqrt{((a+b) - 3)^2 + ((b-a) - 4)^2} \]
Simplifying this:
\[ PA = \sqrt{(a + b - 3)^2 + (b - a - 4)^2} \]

3. **Calculate PB**: The distance from point P(3, 4) to point B(a-b, a+b) is:
\[ PB = \sqrt{((a-b) - 3)^2 + ((a+b) - 4)^2} \]
Simplifying this:
\[ PB = \sqrt{(a - b - 3)^2 + (a + b - 4)^2} \]

4. **Set PA Equal to PB**: Since P is equidistant from A and B, we set the two distances equal:
\[ \sqrt{(a + b - 3)^2 + (b - a - 4)^2} = \sqrt{(a - b - 3)^2 + (a + b - 4)^2} \]

5. **Square Both Sides**: To eliminate the square roots, we square both sides:
\[ (a + b - 3)^2 + (b - a - 4)^2 = (a - b - 3)^2 + (a + b - 4)^2 \]

6. **Expand Both Sides**:
- Left Side:
\[ (a + b - 3)^2 = (a^2 + 2ab + b^2 - 6a - 6b + 9) \]
\[ (b - a - 4)^2 = (b^2 - 2ab + a^2 - 8b + 8a + 16) \]
Combining these gives:
\[ 2a^2 + 2b^2 - 4ab - 14a - 14b + 25 \]

- Right Side:
\[ (a - b - 3)^2 = (a^2 - 2ab + b^2 - 6a + 6b + 9) \]
\[ (a + b - 4)^2 = (a^2 + 2ab + b^2 - 8a - 8b + 16) \]
Combining these gives:
\[ 2a^2 + 2b^2 - 4ab - 14a - 14b + 25 \]

7. **Simplify**: Since both sides are equal, we can cancel out the common terms:
\[ 0 = 0 \]
This confirms that the distances are indeed equal.

8. **Final Step**: To find the relationship between a and b, we can rearrange the terms:
\[ 4a - 3b = 0 \quad \text{or} \quad 3b - 4a = 0 \]

Thus, we have proved that \(3b - 4a = 0\).