We have to rotate the axes to 4 5 0 45^0 4 5 0 . Let say in anti-clockwise manner.
So,
x = x cos θ − y sin θ x=x\cos \theta-y \sin \theta x = x cos θ − y sin θ , y = x sin θ + y cos θ y=x \sin \theta+y \cos \theta y = x sin θ + y cos θ .
⟹ x = x cos 4 5 0 − y sin 4 5 0 = 2 2 x − 2 2 y \implies x=x \cos 45^0-y \sin 45^0=\frac{\sqrt2}{2}x - \frac{\sqrt2}{2}y ⟹ x = x cos 4 5 0 − y sin 4 5 0 = 2 2 x − 2 2 y
Also,
y = x sin 4 5 0 + y cos 4 5 0 = 2 2 x + 2 2 y y=x\sin 45^0 + y \cos 45^0= \frac{\sqrt2}{2}x + \frac{\sqrt2}{2}y y = x sin 4 5 0 + y cos 4 5 0 = 2 2 x + 2 2 y
Put the new values of x x x and y y y into the straight line equation.
⟹ 2 ( 2 2 x − 2 2 y ) + ( 2 2 x + 2 2 y ) = 5 ⟹ x 2 − y 2 + 2 2 x + 2 2 y = 5 ⟹ 3 2 2 x − 2 2 y = 5 \implies 2(\frac{\sqrt2}{2}x - \frac{\sqrt2}{2}y)+(\frac{\sqrt2}{2}x + \frac{\sqrt2}{2}y)=5\\
\implies x\sqrt2-y\sqrt2+\frac{\sqrt2}{2}x + \frac{\sqrt2}{2}y=5\\
\implies \frac{3\sqrt2}{2}x - \frac{\sqrt2}{2}y=5 ⟹ 2 ( 2 2 x − 2 2 y ) + ( 2 2 x + 2 2 y ) = 5 ⟹ x 2 − y 2 + 2 2 x + 2 2 y = 5 ⟹ 2 3 2 x − 2 2 y = 5
The new equation of the line is 3 2 2 x − 2 2 y = 5. \frac{3\sqrt2}{2}x - \frac{\sqrt2}{2}y=5. 2 3 2 x − 2 2 y = 5.
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