Kindly solve this question from coordinate geometry. [closed]

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The co-ordinates of a point P referred to a rectangular co-ordinate system where O is the origin are $(1,-2)$. The axes are rotated about 0 through angle theta, if coordinates of the new P are $(k-1,k+1)$, then $k^2$?




geometry analytic-geometry




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closed as off-topic by Morgan Rodgers, Arnaud D., amWhy, José Carlos Santos, Nosrati Aug 17 at 17:34


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Morgan Rodgers, Arnaud D., amWhy, José Carlos Santos, Nosrati
If this question can be reworded to fit the rules in the help center, please edit the question.








  • 3




    Welcome to Math SE. What have you tried? Where are you stuck? What tools can you use?
    – mfl
    Aug 16 at 11:14







  • 1




    well i tried rotating the axes, the answer doesn't seem to match, i think it is something related to rectangular axes. And Btw, the answer is 3/2.
    – Abdullah
    Aug 16 at 11:18






  • 1




    If you rotate the axis, does it change the lenght of the vector?
    – mfl
    Aug 16 at 11:19






  • 2




    I am telling you my idea. Under a rotation with respect to the origin the length of the vector doesn't change. So the length of $(1,-2)$ is the same as the lenth of $(k-1,k+1).$
    – mfl
    Aug 16 at 11:23







  • 3




    You're welcome. It was a pleasure. And, please, for further questions write your work/ideas in the body question.
    – mfl
    Aug 16 at 11:29















up vote
-1
down vote

favorite












The co-ordinates of a point P referred to a rectangular co-ordinate system where O is the origin are $(1,-2)$. The axes are rotated about 0 through angle theta, if coordinates of the new P are $(k-1,k+1)$, then $k^2$?




geometry analytic-geometry




share|cite|improve this question














closed as off-topic by Morgan Rodgers, Arnaud D., amWhy, José Carlos Santos, Nosrati Aug 17 at 17:34


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Morgan Rodgers, Arnaud D., amWhy, José Carlos Santos, Nosrati
If this question can be reworded to fit the rules in the help center, please edit the question.








  • 3




    Welcome to Math SE. What have you tried? Where are you stuck? What tools can you use?
    – mfl
    Aug 16 at 11:14







  • 1




    well i tried rotating the axes, the answer doesn't seem to match, i think it is something related to rectangular axes. And Btw, the answer is 3/2.
    – Abdullah
    Aug 16 at 11:18






  • 1




    If you rotate the axis, does it change the lenght of the vector?
    – mfl
    Aug 16 at 11:19






  • 2




    I am telling you my idea. Under a rotation with respect to the origin the length of the vector doesn't change. So the length of $(1,-2)$ is the same as the lenth of $(k-1,k+1).$
    – mfl
    Aug 16 at 11:23







  • 3




    You're welcome. It was a pleasure. And, please, for further questions write your work/ideas in the body question.
    – mfl
    Aug 16 at 11:29













up vote
-1
down vote

favorite









up vote
-1
down vote

favorite











The co-ordinates of a point P referred to a rectangular co-ordinate system where O is the origin are $(1,-2)$. The axes are rotated about 0 through angle theta, if coordinates of the new P are $(k-1,k+1)$, then $k^2$?




geometry analytic-geometry




share|cite|improve this question














The co-ordinates of a point P referred to a rectangular co-ordinate system where O is the origin are $(1,-2)$. The axes are rotated about 0 through angle theta, if coordinates of the new P are $(k-1,k+1)$, then $k^2$?





geometry analytic-geometry





share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 16 at 11:19

























asked Aug 16 at 11:12









Abdullah

12




12




closed as off-topic by Morgan Rodgers, Arnaud D., amWhy, José Carlos Santos, Nosrati Aug 17 at 17:34


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Morgan Rodgers, Arnaud D., amWhy, José Carlos Santos, Nosrati
If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by Morgan Rodgers, Arnaud D., amWhy, José Carlos Santos, Nosrati Aug 17 at 17:34


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Morgan Rodgers, Arnaud D., amWhy, José Carlos Santos, Nosrati
If this question can be reworded to fit the rules in the help center, please edit the question.







  • 3




    Welcome to Math SE. What have you tried? Where are you stuck? What tools can you use?
    – mfl
    Aug 16 at 11:14







  • 1




    well i tried rotating the axes, the answer doesn't seem to match, i think it is something related to rectangular axes. And Btw, the answer is 3/2.
    – Abdullah
    Aug 16 at 11:18






  • 1




    If you rotate the axis, does it change the lenght of the vector?
    – mfl
    Aug 16 at 11:19






  • 2




    I am telling you my idea. Under a rotation with respect to the origin the length of the vector doesn't change. So the length of $(1,-2)$ is the same as the lenth of $(k-1,k+1).$
    – mfl
    Aug 16 at 11:23







  • 3




    You're welcome. It was a pleasure. And, please, for further questions write your work/ideas in the body question.
    – mfl
    Aug 16 at 11:29













  • 3




    Welcome to Math SE. What have you tried? Where are you stuck? What tools can you use?
    – mfl
    Aug 16 at 11:14







  • 1




    well i tried rotating the axes, the answer doesn't seem to match, i think it is something related to rectangular axes. And Btw, the answer is 3/2.
    – Abdullah
    Aug 16 at 11:18






  • 1




    If you rotate the axis, does it change the lenght of the vector?
    – mfl
    Aug 16 at 11:19






  • 2




    I am telling you my idea. Under a rotation with respect to the origin the length of the vector doesn't change. So the length of $(1,-2)$ is the same as the lenth of $(k-1,k+1).$
    – mfl
    Aug 16 at 11:23







  • 3




    You're welcome. It was a pleasure. And, please, for further questions write your work/ideas in the body question.
    – mfl
    Aug 16 at 11:29








3




3




Welcome to Math SE. What have you tried? Where are you stuck? What tools can you use?
– mfl
Aug 16 at 11:14





Welcome to Math SE. What have you tried? Where are you stuck? What tools can you use?
– mfl
Aug 16 at 11:14





1




1




well i tried rotating the axes, the answer doesn't seem to match, i think it is something related to rectangular axes. And Btw, the answer is 3/2.
– Abdullah
Aug 16 at 11:18




well i tried rotating the axes, the answer doesn't seem to match, i think it is something related to rectangular axes. And Btw, the answer is 3/2.
– Abdullah
Aug 16 at 11:18




1




1




If you rotate the axis, does it change the lenght of the vector?
– mfl
Aug 16 at 11:19




If you rotate the axis, does it change the lenght of the vector?
– mfl
Aug 16 at 11:19




2




2




I am telling you my idea. Under a rotation with respect to the origin the length of the vector doesn't change. So the length of $(1,-2)$ is the same as the lenth of $(k-1,k+1).$
– mfl
Aug 16 at 11:23





I am telling you my idea. Under a rotation with respect to the origin the length of the vector doesn't change. So the length of $(1,-2)$ is the same as the lenth of $(k-1,k+1).$
– mfl
Aug 16 at 11:23





3




3




You're welcome. It was a pleasure. And, please, for further questions write your work/ideas in the body question.
– mfl
Aug 16 at 11:29





You're welcome. It was a pleasure. And, please, for further questions write your work/ideas in the body question.
– mfl
Aug 16 at 11:29











1 Answer
1






active

oldest

votes

















up vote
-1
down vote



accepted










Rotations preserve length and because it's a rotation with respect to the orgin, the distance from $(1,-2)$ to the orgin is equal to the distance from $(k-1, k+1)$ to the orgin. We can write the equation $$sqrt1^2+(-2)^2=sqrt(k-1)^2+(k+1)^2.$$
Squaring both sides and expanding gives us $$5=2k^2+2.$$
Thus, $$2k^2=3$$ and $$boxedk^2=frac32.$$






share|cite|improve this answer




















  • Thanks, btw if rotation is not with respect to the origin, will the length still be constant?
    – Abdullah
    Aug 16 at 19:52










  • Nope, the distance to the orgin will not be constant if it's rotated with respect to any other point.
    – thejudge333
    Aug 17 at 1:42










  • I understand that the distance to the origin to the origin won't be constant as such, but i meant to ask about the 'vector length'
    – Abdullah
    Aug 17 at 10:40

















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
-1
down vote



accepted










Rotations preserve length and because it's a rotation with respect to the orgin, the distance from $(1,-2)$ to the orgin is equal to the distance from $(k-1, k+1)$ to the orgin. We can write the equation $$sqrt1^2+(-2)^2=sqrt(k-1)^2+(k+1)^2.$$
Squaring both sides and expanding gives us $$5=2k^2+2.$$
Thus, $$2k^2=3$$ and $$boxedk^2=frac32.$$






share|cite|improve this answer




















  • Thanks, btw if rotation is not with respect to the origin, will the length still be constant?
    – Abdullah
    Aug 16 at 19:52










  • Nope, the distance to the orgin will not be constant if it's rotated with respect to any other point.
    – thejudge333
    Aug 17 at 1:42










  • I understand that the distance to the origin to the origin won't be constant as such, but i meant to ask about the 'vector length'
    – Abdullah
    Aug 17 at 10:40














up vote
-1
down vote



accepted










Rotations preserve length and because it's a rotation with respect to the orgin, the distance from $(1,-2)$ to the orgin is equal to the distance from $(k-1, k+1)$ to the orgin. We can write the equation $$sqrt1^2+(-2)^2=sqrt(k-1)^2+(k+1)^2.$$
Squaring both sides and expanding gives us $$5=2k^2+2.$$
Thus, $$2k^2=3$$ and $$boxedk^2=frac32.$$






share|cite|improve this answer




















  • Thanks, btw if rotation is not with respect to the origin, will the length still be constant?
    – Abdullah
    Aug 16 at 19:52










  • Nope, the distance to the orgin will not be constant if it's rotated with respect to any other point.
    – thejudge333
    Aug 17 at 1:42










  • I understand that the distance to the origin to the origin won't be constant as such, but i meant to ask about the 'vector length'
    – Abdullah
    Aug 17 at 10:40












up vote
-1
down vote



accepted







up vote
-1
down vote



accepted






Rotations preserve length and because it's a rotation with respect to the orgin, the distance from $(1,-2)$ to the orgin is equal to the distance from $(k-1, k+1)$ to the orgin. We can write the equation $$sqrt1^2+(-2)^2=sqrt(k-1)^2+(k+1)^2.$$
Squaring both sides and expanding gives us $$5=2k^2+2.$$
Thus, $$2k^2=3$$ and $$boxedk^2=frac32.$$






share|cite|improve this answer












Rotations preserve length and because it's a rotation with respect to the orgin, the distance from $(1,-2)$ to the orgin is equal to the distance from $(k-1, k+1)$ to the orgin. We can write the equation $$sqrt1^2+(-2)^2=sqrt(k-1)^2+(k+1)^2.$$
Squaring both sides and expanding gives us $$5=2k^2+2.$$
Thus, $$2k^2=3$$ and $$boxedk^2=frac32.$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Aug 16 at 15:37









thejudge333

186




186











  • Thanks, btw if rotation is not with respect to the origin, will the length still be constant?
    – Abdullah
    Aug 16 at 19:52










  • Nope, the distance to the orgin will not be constant if it's rotated with respect to any other point.
    – thejudge333
    Aug 17 at 1:42










  • I understand that the distance to the origin to the origin won't be constant as such, but i meant to ask about the 'vector length'
    – Abdullah
    Aug 17 at 10:40
















  • Thanks, btw if rotation is not with respect to the origin, will the length still be constant?
    – Abdullah
    Aug 16 at 19:52










  • Nope, the distance to the orgin will not be constant if it's rotated with respect to any other point.
    – thejudge333
    Aug 17 at 1:42










  • I understand that the distance to the origin to the origin won't be constant as such, but i meant to ask about the 'vector length'
    – Abdullah
    Aug 17 at 10:40















Thanks, btw if rotation is not with respect to the origin, will the length still be constant?
– Abdullah
Aug 16 at 19:52




Thanks, btw if rotation is not with respect to the origin, will the length still be constant?
– Abdullah
Aug 16 at 19:52












Nope, the distance to the orgin will not be constant if it's rotated with respect to any other point.
– thejudge333
Aug 17 at 1:42




Nope, the distance to the orgin will not be constant if it's rotated with respect to any other point.
– thejudge333
Aug 17 at 1:42












I understand that the distance to the origin to the origin won't be constant as such, but i meant to ask about the 'vector length'
– Abdullah
Aug 17 at 10:40




I understand that the distance to the origin to the origin won't be constant as such, but i meant to ask about the 'vector length'
– Abdullah
Aug 17 at 10:40


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