April 1st, 2019, 07:46 PM  #21 
Global Moderator Joined: Dec 2006 Posts: 21,105 Thanks: 2324  Can you clarify this by giving the coordinates you used for H, S and M, your calculated midpoints and slopes, etc? (Your previous posts aren't entirely consistent.)

April 1st, 2019, 09:41 PM  #22 
Senior Member Joined: Mar 2019 From: TTF area Posts: 177 Thanks: 1 
The midpoint of HS = (63.5,25) The midpoint of MH = (72.5,29) Gradient of HS = 1/14 Gradient of MH = 5/2 Equation of HS is y = 1/14x + 831/28 The equation of MH is y = 5/2x  609/4 making each equation = each other: 1/14x + 831/28 = 5/2x  609/4 : x = 71 putting x into an equation I get; 1/14×(times)71 + 831/28 = 25 y = 25 note / is the fraction sign. Not division sorry about that. coordinates I used: M (80,26) H(65,32) S(66,18) When I draw a graph of these midpoints, these intersect at (71,25). Last edited by skipjack; April 2nd, 2019 at 12:00 AM. 
April 1st, 2019, 09:50 PM  #23 
Senior Member Joined: Mar 2019 From: TTF area Posts: 177 Thanks: 1  
April 1st, 2019, 11:51 PM  #24 
Global Moderator Joined: Dec 2006 Posts: 21,105 Thanks: 2324  If your coordinates are correct, the midpoint of HS = (65.5, 25). The equation of the perpendicular bisector of HS is y = x/14 + 569/28 The equation of the perpendicular bisector of MH is y = 5/2x  609/4 making each equation = each other: 1/14x + 569/28 = 5/2x  609/4 : x = 1208/17 = 71.0588... (and so y = 1727/68 = 25.397...) 
April 2nd, 2019, 12:50 AM  #25 
Senior Member Joined: Mar 2019 From: TTF area Posts: 177 Thanks: 1 
Your right, is my answer acceptable? I did make a typo with 63.5 meant to be 65.5 can I round the answer? Last edited by helpmeddddd; April 2nd, 2019 at 12:57 AM. 
April 2nd, 2019, 11:25 AM  #26  
Senior Member Joined: Mar 2019 From: TTF area Posts: 177 Thanks: 1  Quote:
Last edited by skipjack; April 2nd, 2019 at 12:42 PM. Reason: to change final "," to "."  
April 2nd, 2019, 12:10 PM  #27 
Senior Member Joined: Mar 2019 From: TTF area Posts: 177 Thanks: 1  
April 2nd, 2019, 12:47 PM  #28 
Global Moderator Joined: Dec 2006 Posts: 21,105 Thanks: 2324 
Whilst rounding is usually acceptable, skeeter was using the perpendicular bisector of MS, whereas you had used the perpendicular bisectors of HS and MH. The problem wasn't explained clearly enough to establish which approach is appropriate. Perhaps neither approach is appropriate.

April 2nd, 2019, 01:00 PM  #29  
Senior Member Joined: Mar 2019 From: TTF area Posts: 177 Thanks: 1  Quote:
Last edited by skipjack; April 2nd, 2019 at 02:01 PM.  

Tags 
geometry, question, time 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
time question...PLEASE HELP  ron246  Trigonometry  1  May 20th, 2014 10:07 PM 
Time / Percentage Question  frogboxer  Elementary Math  1  July 8th, 2013 08:20 AM 
Probably the simplest question of all time.  sterces  Elementary Math  4  September 7th, 2010 12:53 PM 
physics time question  Darpa.Chief  Physics  1  April 16th, 2010 01:57 PM 
HELLO, anybody have time to do statistics question or....  peterle1  Probability and Statistics  1  February 27th, 2010 06:02 PM 