What is the image of P(7, -5) reflected across the line #y=2#?

1 Answer
Jan 25, 2018

# (7,9)#.

Explanation:

Let, #P'(x',y')# be the image of #P(7,-5)# in the line # l :y=2#.

Then, we know from Geometry, that, #l# is the #bot"-bisector"# of #PP'#.

Therefore, (1) : The mid-point** of #PP'# must lie on #l#.

(2) : #PP' bot l#.

But, the mid-pt. of #PP'# is #((x'+7)/2,(y'-5)/2)#.

# ((x'+7)/2,(y'-5)/2) in l rArr (y'-5)/2=2 rArr y'=9...........(1')#.

Note that, #l# is horizontal, so that, #PP'# must be, because of (2),

vertical.

Recall that, the eqn. of a vertical line through #(x_0,y_0)# is, #x=x_0#.

Hence, the eqn. of #PP',# which passes through #P(7,-5)# is,

# PP' : x=7#.

#P'(x',y') in PP' rArr x'=7.............................................(2')#.

Thus, by #(1') and (2'),# the reqd. image point is #P'=P'(7,9)#.