The point P(1,4) occupies the positions A, B | vedclass

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

Question

(B) Reflection of point $P(1,4)$ about the line $y=x$ is $A(4,1)$.After translation of point $A(4,1)$ through a distance of $1$ unit along the positiv

Vedclass · Accepted Answer

$(2 \sqrt{2}, 3 \sqrt{2})$

Vedclass · Answer

(B) Reflection of point $P(1,4)$ about the line $y=x$ is $A(4,1)$.After translation of point $A(4,1)$ through a distance of $1$ unit along the positive direction of $X$-axis,the new coordinate is $B(5,1)$.After rotation of the point $B(5,1)$ through an angle $	heta = \frac{\pi}{4}$ about the origin in the anti-clockwise direction,the new coordinates $(x', y')$ are given by:$x' = x \cos 	heta - y \sin 	heta = 5 \cos \frac{\pi}{4} - 1 \sin \frac{\pi}{4} = \frac{5}{\sqrt{2}} - \frac{1}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}$.$y' = x \sin 	heta + y \cos 	heta = 5 \sin \frac{\pi}{4} + 1 \cos \frac{\pi}{4} = \frac{5}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{6}{\sqrt{2}} = 3\sqrt{2}$.Thus,the coordinates of $C$ are $(2\sqrt{2}, 3\sqrt{2})$.Hence,option $B$ is correct.

Similar Questions

The transformed equation of the curve $2x^2+y^2-3x+5y-8=0$ when the origin is translated to the point $(-1, 2)$ is

If the origin is shifted to the point $(-2, 1)$,what are the new coordinates of the point $(4, -5)$?

When the origin is shifted to $(2, 3)$,the transformed equation of a curve becomes $x^2+3xy-2y^2+17x-7y-11=0$. Find the original equation of the curve.

Transforming to parallel axes through a point $(p, q)$, the equation $2x^2 + 3xy + 4y^2 + x + 18y + 25 = 0$ becomes $2x^2 + 3xy + 4y^2 = 1$. Then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module