April 1st, 2019, 06:46 PM  #21 
Global Moderator Joined: Dec 2006 Posts: 20,746 Thanks: 2133  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, 08:41 PM  #22 
Member Joined: Mar 2019 From: TTF area Posts: 83 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 1st, 2019 at 11:00 PM. 
April 1st, 2019, 08:50 PM  #23 
Member Joined: Mar 2019 From: TTF area Posts: 83 Thanks: 1  
April 1st, 2019, 10:51 PM  #24 
Global Moderator Joined: Dec 2006 Posts: 20,746 Thanks: 2133  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 1st, 2019, 11:50 PM  #25 
Member Joined: Mar 2019 From: TTF area Posts: 83 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 1st, 2019 at 11:57 PM. 
April 2nd, 2019, 10:25 AM  #26  
Member Joined: Mar 2019 From: TTF area Posts: 83 Thanks: 1  Quote:
Last edited by skipjack; April 2nd, 2019 at 11:42 AM. Reason: to change final "," to "."  
April 2nd, 2019, 11:10 AM  #27 
Member Joined: Mar 2019 From: TTF area Posts: 83 Thanks: 1  
April 2nd, 2019, 11:47 AM  #28 
Global Moderator Joined: Dec 2006 Posts: 20,746 Thanks: 2133 
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, 12:00 PM  #29  
Member Joined: Mar 2019 From: TTF area Posts: 83 Thanks: 1  Quote:
Last edited by skipjack; April 2nd, 2019 at 01:01 PM.  

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