The distance between (2p – 1, 3q + 2) and (p + 3, 2q – 1) is same as the distance between (p, q) and (x, y) where (x, y) =

1. Break Down the Problem

We need to find coordinates $(x, y)$ such that the distance between the points $(2p - 1, 3q + 2)$ and $(p + 3, 2q - 1)$ is equal to the distance between $(p, q)$ and $(x, y)$.

2. Relevant Concepts

The distance $d$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

3. Analysis and Detail

1. Calculate the distance between $(2p - 1, 3q + 2)$ and $(p + 3, 2q - 1)$:

   • Let:
     • $(x_1, y_1) = (2p - 1, 3q + 2)$
     • $(x_2, y_2) = (p + 3, 2q - 1)$
   • The distance $d_1$ is: $d_1 = \sqrt{((p + 3) - (2p - 1))^2 + ((2q - 1) - (3q + 2))^2}$
   • Simplifying: $d_1 = \sqrt{(p + 3 - 2p + 1)^2 + (2q - 1 - 3q - 2)^2}$ $d_1 = \sqrt{(-p + 4)^2 + (-q - 3)^2}$ $d_1 = \sqrt{(p - 4)^2 + (q + 3)^2}$

2. Let the distance between $(p, q)$ and $(x, y)$ be $d_2$:

   • The distance $d_2$ is: $d_2 = \sqrt{(x - p)^2 + (y - q)^2}$

3. Set the distances equal: $\sqrt{(p - 4)^2 + (q + 3)^2} = \sqrt{(x - p)^2 + (y - q)^2}$

   • Squaring both sides gives: $(p - 4)^2 + (q + 3)^2 = (x - p)^2 + (y - q)^2$

4. Verify and Summarize

To find $(x, y)$ that satisfies this, we observe:

This equation represents a geometric circle centered at $(4, -3)$ with radius equal to the distance from $(p, q)$ to $(4, -3)$. The coordinates $(x, y)$ lie on this circle.

Final Answer

Thus, the coordinates $(x, y)$ satisfy the equation: $(x - p)^2 + (y - q)^2 = (p - 4)^2 + (q + 3)^2$ This describes a locus of points $(x, y)$ for any given $(p, q)$.

To find specific coordinates $(x, y)$ would require additional information or constraints.