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Coordinate problm

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by gmatclubmember » Sun Sep 11, 2011 8:05 am
jogi1984 wrote:Give mirror image of (-3,2) across X=Y line ???
I guess the answer is (2,-3). But I calculated it using 2 equations in x and y variables which took me almost like 2 mins. Someone can please come up with a short cut? What if the mirror is not line x=y but ax+by+c=0.

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by ravirajsitaram » Sun Sep 11, 2011 8:53 am
Straight line -- ax+by+c = 0
point (x1,y1)

To find mirror image point :
A straight line with x=x1, touches the line ax+by+c at point (x1,(-c-ax1)/b)
similarly line y=y1 touches the line at ((-c-by1)/a,y1).
Now the mirror image of (x1,y1) will be equidistant wrt line in ((-c-by1)/a,(-c-ax1)/b).
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by mehrasa » Sun Sep 11, 2011 8:55 am
I prefer to explain 3 different situation under this topic

when we want to find reflected point based on y=x we just change the position of x-coordinate and y-coordinate (e.g. (2,3) --> (3,2))
as for x-axis => the x-coordinate will not change while y-coordinate sign reverse (e.g (3.2) --> (3,-2)
as for y-axis ==> the y-coordinate remain the same and x-coordinate sign change (3,2) --> (-3,2)
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by sl750 » Sun Sep 11, 2011 9:07 am
Slope of y=x is 1. The mirror image of (-3,2) along the y axis it is (3,2). Along the x-axis it is (-3,-2)
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