SOLUTION: find the point whose distance from (7,-3) is square root of 58 and whose abscissa equals its ordinate. Please show your solution.
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Question 864936: find the point whose distance from (7,-3) is square root of 58 and whose abscissa equals its ordinate. Please show your solution.
Answer by richwmiller(17219) (Show Source): You can put this solution on YOUR website!
the point will be (y,y) or (x,x) since the abscissa = the ordinate x=y
sqrt(58)
58=(7-x)^2+(-3-x)^2
(0,0) and (4,4) are sqrt(58) from (7,-3)
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