=0, we get 4x−y−3=0‌ and ‌−x−2y+3=0 Solving these two equations for x=h and y=k 4h−k−3‌=0‌ and ‌−h−2k+3=0 ∴h=1,k‌=1 So, the new origin is (h,k)=(1,1) We know that a general second-degree equation Ax2+Bxy+Cy2+Dx+Ey+F=0, the angle of rotation θ satisfies tan‌2‌θ=‌
B
A−C
Here, S≡2x2−xy−y2−3x+3y=0, we have ‌A=2B=−1‌ and ‌C=−1 ‌∴tan‌2‌θ=‌