If the axes are rotated through an angle of 45∘ and the point p has new co-ordinates (22,2) then original coordinates of p are

Given θ=45ο,Use x=Xcos⁡θ−Ysin⁡θy=Xsin⁡θ+Ycos⁡θ ⇒x=22(12)-2(12)=1 y=22(12)+2(12)=3

If the axes are rotated through an angle of 45∘ and the point p has new co-ordinates (22,2) then original coordinates of p are

Given θ=45ο,Use x=Xcos⁡θ−Ysin⁡θy=Xsin⁡θ+Ycos⁡θ ⇒x=22(12)-2(12)=1 y=22(12)+2(12)=3

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

If the axes are rotated through an angle of $45^{\circ}$ and the point p has new co-ordinates $(2 \sqrt{2}, \sqrt{2})$ then original coordinates of p are

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

(3,3)

(-1,3)

(3,1)

(1,3)

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$G i v e n θ = 45^{ο}, U s e \begin{aligned} x = X \cos θ - Y \sin θ \\ y = X \sin θ + Y \cos θ \end{aligned} \Rightarrow x = 2 \sqrt{2} (\frac{1}{\sqrt{2}}) - \sqrt{2} (\frac{1}{\sqrt{2}}) = 1 y = 2 \sqrt{2} (\frac{1}{\sqrt{2}}) + \sqrt{2} (\frac{1}{\sqrt{2}}) = 3$