Let L be the line 2x + y = 2. If the axes are | vedclass

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(C) The equation of the line is $2x + y = 2$,which can be written in intercept form as $\frac{x}{1} + \frac{y}{2} = 1$. Here,the intercepts on the ori

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$2\sqrt{2}/3$ and $-2\sqrt{2}$

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(C) The equation of the line is $2x + y = 2$,which can be written in intercept form as $\frac{x}{1} + \frac{y}{2} = 1$. Here,the intercepts on the original axes are $a = 1$ and $b = 2$.When the axes are rotated by an angle $\alpha = 45^\circ$ in the anticlockwise direction,the new intercepts $p$ and $q$ are given by the formulas:$\frac{1}{p} = \frac{1}{a} \cos \alpha + \frac{1}{b} \sin \alpha$$\frac{1}{q} = -\frac{1}{a} \sin \alpha + \frac{1}{b} \cos \alpha$Substituting $a = 1$,$b = 2$,and $\alpha = 45^\circ$:$\frac{1}{p} = \frac{1}{1} \cdot \frac{1}{\sqrt{2}} + \frac{1}{2} \cdot \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} + \frac{1}{2\sqrt{2}} = \frac{3}{2\sqrt{2}}$ $\Rightarrow p = \frac{2\sqrt{2}}{3}$$\frac{1}{q} = -\frac{1}{1} \cdot \frac{1}{\sqrt{2}} + \frac{1}{2} \cdot \frac{1}{\sqrt{2}} = -\frac{1}{\sqrt{2}} + \frac{1}{2\sqrt{2}} = -\frac{1}{2\sqrt{2}}$ $\Rightarrow q = -2\sqrt{2}$Thus,the intercepts are $\frac{2\sqrt{2}}{3}$ and $-2\sqrt{2}$.

Let $L$ be the line $2x + y = 2$. If the axes are rotated by $45^\circ$,then the intercepts made by the line $L$ on the new axes are respectively

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