The co-ordinate axes are rotated through an angle $135^{\circ}$. If the co-ordinates of a point $P$ in the…

The co-ordinate axes are rotated through an angle $135^{\circ}$. If the co-ordinates of a point $P$ in the new system are known to be $(4,-3)$, then the co-ordinates of $P$ in the original system are
  1. $\left(\frac{1}{\sqrt{2}}, \frac{7}{\sqrt{2}}\right)$
  2. $\left(\frac{1}{\sqrt{2}},-\frac{7}{\sqrt{2}}\right)$
  3. $\left(-\frac{1}{\sqrt{2}},-\frac{7}{\sqrt{2}}\right)$
  4. $\left(-\frac{1}{\sqrt{2}}, \frac{7}{\sqrt{2}}\right)$

Solution

Let $\left(x_1, y_1\right)$ be the co-ordinate of new system, then $x_1=x \cos \theta+y \sin \theta$ $y_1=-x \sin \theta+y \cos \theta$ Given that $\left(x_1, y_1\right)=(4,-3)$ and $\theta=135^{\circ}$ $4=x \cos 135^{\circ}+y \sin 135^{\circ}$ $\Rightarrow \quad 4=-\frac{x}{\sqrt{2}}+\frac{y}{\sqrt{2}}$\ldots(\mathrm{i})$ $-3=-x \sin 135^{\circ}+y \cos 135^{\circ}$ $\Rightarrow \quad-3=-\frac{x}{\sqrt{2}}-\frac{y}{\sqrt{2}}$\ldots(\mathrm{ii})$ On adding Eqs. (i) and (ii), we get $1=-\frac{2 x}{\sqrt{2}}$ $\Rightarrow \quad x=-\frac{1}{\sqrt{2}}$ On subtracting Eqs. (i) and (ii), we get $7=\frac{2 y}{\sqrt{2}}$ $\Rightarrow \quad y=\frac{7}{\sqrt{2}}$ Thus $(x, y)=\left(-\frac{1}{\sqrt{2}}, \frac{7}{\sqrt{2}}\right)$

Asked in: AP EAMCET 2003

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