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14) reflect across the line y = xM(3,6) ► M'N(5, 10) ► NP(2,-5) →P

Sagot :

Problem

14) reflect across the line y = x

M(3,6) ► M'

N(5, 10) ► N

P(2,-5) →P

Solution

For this case we need to remember that if we have a point (x,y) when we reflect respect to the line y=x the new coordinates are (y,x)

And using this formula we have:

M' = (6,3)

N'= (10,5)

P' =(-5,2)

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