The locus of a point equidistant from two points p(6,2) and R(4,2) is a perpendicular bisector of PR passing through?

Talented007

1 year ago

i can't view explanation it keeps showing temporary error

Nathan_X

1 year ago

unsanitized build exception detected

edutara

2 months ago

Formula for midpoint=(x1 + x2)/2 (y1 + y2)/2
(6 + 4)/2 (2+2)/2 =10/2 4/2 = 5,2
Which makes option b the correct answer

jcode12

1 year ago

temporary node crnode encountered

SuccOffem

4 months ago

ok

agbolahan 1765

10 years ago

x2-x1 /2, y2-y1/2 4-6/2, 2-2/2

succy34

1 year ago

I am not seeing the solution
can someone help me

Grais

3 years ago

6+2 can never give 10

Odemosad

1 year ago

Option B. (5,2)