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Mathematics: Post your doubts here!

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Usually for Enlargement/Reflection questions it get to be a shape on the graph as a rectangle, hexagon..etc. whenever we enlarge an image we usually multiply the vertices of the shape by the enlargement Factor or matrix to get the new set of prime points forming a shape.

OKAY what if this time in the external the shape was a circle...
and the question states to enlarge it using a certain factor or matrix for instance by then we really can't just multiply vertices......because it's rounded what shall we do in this case??

Thank you help is appreciated.

the points of the circle will have co-ordinates on the graph, use 2 co-ordinates,for example, one on the extreme left and extreme right of the circle, and get the new position of the points. then draw a circle using these two points with a circle.
 
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the points of the circle will have co-ordinates on the graph, use 2 co-ordinates,for example, one on the extreme left and extreme right of the circle, and get the new position of the points. then draw a circle using these two points with a circle.
Thank you !

reflection in x axis yes that will always be the matrix representing that transformation.
however for refletion in y axis and etc there are different matrices.

For the reflections on the Y-axis there are different matrices depending upon what..?
I mean what kind of different matrices that can possibly be there?..
 
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Thank you !
For the reflections on the Y-axis there are different matrices depending upon what..?
I mean what kind of different matrices that can possibly be there?..

I meant there are different matrices representing transformations depending upon the transformation.
for example matrix representing reflection in x axis is different from matrix representing reflection in y axis.
 
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Thank you so much bro!
One more 10 a in same paper please
So basically the diameter cuts the tangent into 2 equal parts..
and the radius bisencts the angle and we know that each angle in am equilateral triangle is 60
 

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Question is saying it's boundry is equidistant from Ps and Pq so yes boundry is touching the equidistant line
 
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