Answer :

Given, equation x2 + xy – 3x + 2 = 0


Let’s assume that the origin is shifted at point (p, q).


To find: The shifted point (p, q) satisfying the question’s conditions.


We know that, when we transform origin from (0, 0) to an arbitrary point (p, q), the new coordinates for the point (x, y) becomes (x + p, y + q), and hence an equation with two variables x and y must be transformed accordingly replacing x with x + p, and y with y + q in original equation.


Since, origin has been shifted from (0, 0) to (p, q); therefore any arbitrary point (x, y) will also be converted as (x + p, y + q).


The New equation hence becomes:


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


= 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.


Solving these simultaneous equations we have p = 1 and q = 1 from first equation. And, p = 1 and q = 1 satisfies p2 + pq – 3p - q + 2 = 0.


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


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