**Question:**

Find what the given equation becomes when the origin is shifted to the point (1, 1).

$x^{2}+x y-3 x-y+2=0$

**Solution:**

Let the new origin be (h, k) = (1, 1)

Then, the transformation formula become:

x = X + 1 and y = Y + 1

Substituting the value of x and y in the given equation, we get

$x^{2}+x y-3 x-y+2=0$

Thus

$(X+1)^{2}+(X+1)(Y+1)-3(X+1)-(Y+1)+2=0$

$\Rightarrow\left(X^{2}+1+2 X\right)+X Y+X+Y+1-3 X-3-Y-1+2=0$

$\Rightarrow X^{2}+1+2 X+X Y-2 X-1=0$

$\Rightarrow X^{2}+X Y=0$

Hence, the transformed equation is $X^{2}+X Y=0$