SOLUTION: The midpoint of MN is point P at (–4, 6). If point M is at (8, –2), what are the coordinates of point N? (–16, 14) (–10, 5) (2, 2) (6, 2)

Algebra ->  Points-lines-and-rays -> SOLUTION: The midpoint of MN is point P at (–4, 6). If point M is at (8, –2), what are the coordinates of point N? (–16, 14) (–10, 5) (2, 2) (6, 2)      Log On


   



Question 989546: The midpoint of MN is point P at (–4, 6). If point M is at (8, –2), what are the coordinates of point N?
(–16, 14)
(–10, 5)
(2, 2)
(6, 2)

Found 2 solutions by solver91311, josmiceli:
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


Use the midpoint formulas. Substitute your values for and then solve for and , the coordinates of the other endpoint.

and



John

My calculator said it, I believe it, that settles it

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
The x-coordinate of the midpoint is at:
+x%5Bp%5D+=+%28+x%5B1%5D+%2B+x%5B2%5D+%29+%2F+2+
where +x%5B1%5D+ and +x%5B2%5D+ are
x-coordinates of the endpoints
----------------------------------
The y-coordinate of the midpoint is at:
+y%5Bp%5D+=+%28+y%5B1%5D+%2B+y%5B2%5D+%29+%2F+2+
where +y%5B1%5D+ and +y%5B2%5D+ are
y-coordinates of the endpoints
----------------------------------
I will say that:
( x[1], y[1] ) = M( 8, -2 )
( x[p], y[p] ) = P( -4, 6 )
I need to find N( x[2], y[2] )
--------------------------
First, using:
+x%5Bp%5D+=+%28+x%5B1%5D+%2B+x%5B2%5D+%29+%2F+2+
+-4+=+%28+8+%2B+x%5B2%5D+%29+%2F+2+
+-8+=+8+%2B+x%5B2%5D+
+x%5B2%5D+=+-16+
--------------------
Next, using:
+y%5Bp%5D+=+%28+y%5B1%5D+%2B+y%5B2%5D+%29+%2F+2+
+6+=+%28+-2+%2B+y%5B2%5D+%29+%2F+2+
+12+=+-2+%2B+y%5B2%5D+
+y%5B2%5D+=+14+
--------------------------------
The other endpoint is at: N( -16, 14 )
--------------------------------
check:
+-4+=+%28+8+%2B+x%5B2%5D+%29+%2F+2+
+-4+=+%28+8+%2B%28-16%29+%29+%2F+2+
+-4+=+%28-8%29%2F2+
+-4+=+-4+
and
+6+=+%28+-2+%2B+y%5B2%5D+%29+%2F+2+
+6+=+%28+-2+%2B+14+%29+%2F+2+
+6+=+12%2F2+
+6+=+6+
OK