Solution:

We have the equation x² + xy – 3x + 2 = 0

The origin has been relocated from (0, 0) to  (p, q)

As a result, any arbitrary point (x, y) is changed to (x + p, y + q).

Therefore, the new equation is:

(x + p)² + (x + p)(y + q) – 3(x + p) + 2 = 0

Upon simplification further, we get:

x² + p² + 2px + xy + py + qx + pq – 3x – 3p + 2 = 0

x² + xy + x(2p + q – 3) + y(q – 1) + p² + pq – 3p – q + 2 = 0

For no first degree term, we have 2p + q – 3 = 0 and p – 1 = 0, and

For no constant term we have p² + pq – 3p – q + 2 = 0.

We get p = 1 and q = 1 from the first equation by solving these simultaneous equations.

The values p = 1 and q = 1 satisfies p² + pq – 3p – q + 2 = 0.

Hence, (p, q) = (1, 1) is the point to which origin must be shifted.