A) \[\sqrt{2}\]and 1
B) 1 and \[\sqrt{2}\]
C) \[2\sqrt{2}\]and \[2\sqrt{2}/3\]
D) \[2\sqrt{2}/3\]and \[2\sqrt{2}\]
Correct Answer: C
Solution :
Suppose the axes are rotated in the anticlockwise direction through an angle\[{{45}^{o}}\]. To find the equation of L w.r.t the new axis, we replace x by \[x\cos \alpha -y\sin \alpha \] and by\[x\sin \alpha +y\cos \alpha \], so that equation of line w.r.t. new axes is Þ \[1/1(x\cos {{45}^{o}}-y\sin {{45}^{o}})+\frac{1}{2}(x\sin {{45}^{o}}+y\cos {{45}^{o}})=1\] Since, p, q are the intercept made by the line on the coordinate axes. we have on putting (p, 0) and then (0, q) Þ \[\frac{1}{p}=\frac{1}{a}\cos \alpha +\frac{1}{b}\sin \alpha \Rightarrow \frac{1}{q}=-\frac{1}{a}\sin \alpha +\frac{1}{b}\cos \alpha \] Þ \[\frac{1}{p}=\frac{1}{1}\cos {{45}^{o}}+\frac{1}{2}\sin {{45}^{o}}\] \[\Rightarrow \frac{1}{p}=\frac{1}{\sqrt{2}}+\frac{1}{2}.\frac{1}{\sqrt{2}}=\frac{3}{2\sqrt{2}}\] \ \[p=\frac{2\sqrt{2}}{3}\]; \\[\frac{1}{q}=-\frac{1}{1}\sin {{45}^{o}}+\frac{1}{2}\cos {{45}^{o}}\] \[\frac{1}{q}=\frac{-1}{\sqrt{2}}+\frac{1}{2\sqrt{2}}=-\frac{1}{2\sqrt{2}}\,,\,\,\,\therefore q=2\sqrt{2}\] So intercept made by is assume on the new axis \[\left( 2\sqrt{2}/3,\,\,2\sqrt{2} \right)\]. If the rotation is assume in clockwise direction, so intercept made by the line on the new axes would be \[\left( 2\sqrt{2},\,2\sqrt{2}/3 \right)\].
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