The coordinate axes are rotated through an an | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let $(x, y)$ be the coordinates in the original system and $(x', y')$ be the coordinates in the new system after rotation by an angle $	heta = 13

Vedclass · Accepted Answer

$(-\frac{1}{\sqrt{2}}, \frac{7}{\sqrt{2}})$

Vedclass · Answer

(D) Let $(x, y)$ be the coordinates in the original system and $(x', y')$ be the coordinates in the new system after rotation by an angle $	heta = 135^{\circ}$.The transformation equations are:$x' = x \cos 	heta + y \sin 	heta$$y' = -x \sin 	heta + y \cos 	heta$Given $(x', y') = (4, -3)$ and $	heta = 135^{\circ}$,we have:$4 = x \cos 135^{\circ} + y \sin 135^{\circ} = -\frac{x}{\sqrt{2}} + \frac{y}{\sqrt{2}}$$\Rightarrow -x + y = 4\sqrt{2} \quad (i)$$-3 = -x \sin 135^{\circ} + y \cos 135^{\circ} = -\frac{x}{\sqrt{2}} - \frac{y}{\sqrt{2}}$$\Rightarrow x + y = 3\sqrt{2} \quad (ii)$Adding $(i)$ and $(ii)$ gives $2y = 7\sqrt{2} \Rightarrow y = \frac{7}{\sqrt{2}}$.Subtracting $(i)$ from $(ii)$ gives $2x = -\sqrt{2} \Rightarrow x = -\frac{1}{\sqrt{2}}$.Thus,the coordinates in the original system are $(-\frac{1}{\sqrt{2}}, \frac{7}{\sqrt{2}})$.

The coordinate axes are rotated through an angle $135^{\circ}$. If the coordinates of a point $P$ in the new system are known to be $(4, -3)$,then the coordinates of $P$ in the original system are

Similar Questions

When the coordinate axes are rotated through an angle of $45^{\circ}$ about the origin in the positive direction,if the transformed equation of a curve is $17x^2 - 16xy + 17y^2 = 225$,then the original equation of that curve is:

The point to which the origin should be shifted in order to eliminate the $x$ and $y$ terms from the equation $9x^2+4y^2+10x+12y+1=0$ is

If the area of the region bounded by the curves $y=x^2$ and $x=y^2$ is $k$,then the area of the region bounded by the curves $\frac{x+\sqrt{3} y}{2}=\left(\frac{\sqrt{3} x-y}{2}\right)^2$ and $\frac{\sqrt{3} x-y}{2}=\left(\frac{x+\sqrt{3} y}{2}\right)^2$ is:

The transformed equation of $x^2-y^2+2x+4y=0$ when the origin is shifted to the point $(-1, 2)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module