Question

Let L be the line $2x+y=2$. If axes are rotated by ${45}^0$ in anticlockwise direction the intercepts made by line on new axis=?

• A. $2and1$
• B. $1and2$
• C. $2\left(\sqrt{2}\right)$ and $\frac{2\sqrt{2}}{3}$
• D. $2and2$

Answer 1

Answer: (C)

$2\left(\sqrt{2}\right)$ and $\frac{2\sqrt{2}}{3}$

Suppose the axis are rotated in anti-clockwise direction through an angle of $45°$. To find the equation of the line with respect to the new axis, we replace $x$ by $x\cos \alpha -y\sin \alpha$ and by $x\sin \alpha +y\cos \alpha$ so that the equation of the line with respect to the new axis is

$=>\frac{1}{1}\times (x\cos 45°-y\sin 45°)+\frac{1}{2}\times (x\sin 45°+y\cos 45°)=1$

Since, p and q are the intercept made by the line on the co-ordinate axis, we have on putting $(p,0)$ and then $(0,q)$

$=>\frac{1}{p}=\frac{1}{a}\cos \alpha +\frac{1}{b}\sin \alpha \frac{1}{q}=-\frac{1}{a}\sin \alpha +\frac{1}{b}\cos \alpha \frac{1}{p}=\frac{1}{1}\cos 45°+\frac{1}{2}\sin 45°=>\frac{1}{p}=\frac{1}{\sqrt{2}}+\frac{1}{2}\times \frac{1}{\sqrt{2}}=\frac{3}{2\sqrt{2}}=>p=\frac{2\sqrt{2}}{3}\frac{1}{q}=-\frac{1}{1}\sin 45°+\frac{1}{2}\cos 45°=>\frac{1}{q}=-\frac{1}{\sqrt{2}}+\frac{1}{2}\times \frac{1}{\sqrt{2}}=-\frac{1}{2\sqrt{2}}=>q=2\sqrt{2}$

The intercept made by the line on the new axis is $2\sqrt{2}$ and $\frac{2\sqrt{2}}{3}$