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Question
To what point should the origin be shifted so that the equation x2 + xy − 3x − y + 2 = 0 does not contain any first degree term and constant term?
Solution
Let the origin be shifted to (h, k). Then, x = X + h and y = Y + k.
Substituting x = X + h and y = Y + k in the equation x2 + xy − 3x − y + 2 = 0, we get:
\[\left( X + h \right)^2 + \left( X + h \right)\left( Y + k \right) - 3\left( X + h \right) - \left( Y + k \right) + 2 = 0\]
\[ \Rightarrow X^2 + 2hX + h^2 + XY + kX + hY + hk - 3X - 3h - Y - k + 2 = 0\]
\[ \Rightarrow X^2 + XY + X\left( 2h + k - 3 \right) + Y\left( h - 1 \right) + h^2 + hk - 3h - k + 2 = 0\]
For this equation to be free from the first-degree terms and constant term, we must have
\[2h + k - 3 = 0, h - 1 = 0, h^2 + hk - 3h - k + 2 = 0\]
\[ \Rightarrow h = 1, k = 1, h^2 + hk - 3k - h + 2 = 0\]
Also, h =1 and k = 1 satisfy the equation \[h^2 + hk - 3k - h + 2 = 0\]
Hence, the origin should be shifted to the point (1, 1).
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