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?

Answer 1

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:

$$\begin{gathered} {\left(X+h\right)}^2+\left(X+h\right)\left(Y+k\right)-3\left(X+h\right)-\left(Y+k\right)+2=0 \\ \Rightarrow X^2+2hX+h^2+XY+kX+hY+hk-3X-3h-Y-k+2=0 \\ \Rightarrow X^2+XY+X\left(2h+k-3\right)+Y\left(h-1\right)+h^2+hk-3h-k+2=0 \end{gathered}$$

For this equation to be free from the first-degree terms and constant term, we must have

$$\begin{gathered} 2h+k-3=0,h-1=0,h^2+hk-3h-k+2=0 \\ \Rightarrow h=1,k=1,h^2+hk-3k-h+2=0 \end{gathered}$$

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).