My Math Forum A geometry time question

 Geometry Geometry Math Forum

 April 1st, 2019, 03:55 PM #11 Senior Member   Joined: Mar 2019 From: TTF area Posts: 168 Thanks: 1 H is Hamilton, but that's quite irrelevant. I think the plane crashed at (-71,25) rounded; is this correct? Last edited by skipjack; April 1st, 2019 at 04:24 PM.
April 1st, 2019, 04:10 PM   #12
Math Team

Joined: Jul 2011
From: Texas

Posts: 3,044
Thanks: 1627

Quote:
 Originally Posted by helpmeddddd I think the plane crashed at (-71,25) rounded; is this correct?

Quote:
 Originally Posted by helpmeddddd its approx position is at a place equidistant from Miami and San Juan and on altitude from Miami,
a point equidistant from Miami and San Juan is on the perpendicular bisector of line segment MS.

What about the altitude from Miami to ... where?

Last edited by skipjack; April 1st, 2019 at 04:27 PM.

 April 1st, 2019, 04:18 PM #13 Senior Member   Joined: Mar 2019 From: TTF area Posts: 168 Thanks: 1 The altitude part confuses me also. I believe it says on altitude to Miami, so I don't think we have to solve altitude. Perpendicular bisectors are correct; that's how I solved. I found the perpendicular bisector of HS and MH. Last edited by skipjack; April 1st, 2019 at 04:29 PM.
April 1st, 2019, 04:24 PM   #14
Math Team

Joined: Jul 2011
From: Texas

Posts: 3,044
Thanks: 1627

Quote:
 Originally Posted by helpmeddddd The attitude part confuses me also. I believe it says on altitude to Miami, so I don't think we have to solve altitude. Perpendicular bisectors are correct; that's how I solved. I found the perpendicular bisector of HS and MH.
I thought you said point H was irrelevant?

Last edited by skipjack; April 1st, 2019 at 04:29 PM.

 April 1st, 2019, 04:27 PM #15 Senior Member   Joined: Mar 2019 From: TTF area Posts: 168 Thanks: 1 I see how you could get that; I meant what H stands for is irrelevant. Sorry for this confusion. Basically, I found the midpoints of HS and MH, found their slopes/gradients, then did the equations of the gradients. After solving for x, I put x into one of my equations to solve for y. Last edited by skipjack; April 1st, 2019 at 04:57 PM.
April 1st, 2019, 04:55 PM   #16
Global Moderator

Joined: Dec 2006

Posts: 21,036
Thanks: 2273

Quote:
 Originally Posted by helpmeddddd 1 knot = 1.852 km/hr.
More accurately, 1 knot = 1.853184 km/hr.

 April 1st, 2019, 05:06 PM #17 Math Team     Joined: Jul 2011 From: Texas Posts: 3,044 Thanks: 1627 You have three non-collinear points, M(-80,26) H(-65,32) and S(-66,18 ), that form a triangle. The perpendicular bisector of segment MS passes through its midpoint, (-73,22). The slope of MS is -4/7, therefore, the slope of the perpendicular bisector slope 7/4. The equation of the perpendicular bisector of MS is $y = \dfrac{7}{4}(x + 66) + 18$ The altitude from M to HS passes through M(-80,26) and is perpendicular to HS ... The slope of HS is 14, so the altitude to HS has slope -1/14. The equation of the altitude from M to HS is $y = -\dfrac{1}{14}(x + 80) + 26$ The intersection of the altitude from M to HS and the perpendicular bisector of MS can be found by setting the two lines equal. You calculated (-71,25) rounded. Your y-value is good ... your x-coordinate is off according to my calculation.
April 1st, 2019, 05:21 PM   #18
Math Team

Joined: Jul 2011
From: Texas

Posts: 3,044
Thanks: 1627

Quote:
 Originally Posted by skipjack More accurately, 1 knot = 1.853184 km/hr.
Source for that value?

Last edited by skipjack; April 1st, 2019 at 05:31 PM.

April 1st, 2019, 05:54 PM   #19
Global Moderator

Joined: Dec 2006

Posts: 21,036
Thanks: 2273

Quote:
 Originally Posted by skeeter The perpendicular bisector of segment MS passes through its midpoint, (-73,22). The slope of MS is -4/7, therefore, the slope of the perpendicular bisector slope 7/4. The equation of the perpendicular bisector of MS is $y = \dfrac{7}{4}(x + 66) + 18$ ... your x-coordinate is off according to my calculation.
You gave the equation of the perpendicular bisector of MS incorrectly, as you used the coordinates of S instead of the coordinates of the midpoint of MS.

Quote:
 Originally Posted by skeeter Source for that value?
I had used an outdated value of 6080 feet for the nautical mile. The currently used value is exactly 1.852 km, so you were correct.

According to wikipedia, this value was first defined in 1929 (at an international conference held in Monaco). However, the United States adopted this value in 1954 and Britain adopted it in 1970.

April 1st, 2019, 06:25 PM   #20
Math Team

Joined: Jul 2011
From: Texas

Posts: 3,044
Thanks: 1627

Quote:
 Originally Posted by skipjack You gave the equation of the perpendicular bisector of MS incorrectly, as you used the coordinates of S instead of the coordinates of the midpoint of MS.
I certainly did ... good catch.

perpendicular bisector of MS should be

$y = \dfrac{7}{4}(x+73)+22$

making the intersection coordinates approx (-71,25)

so the OP is correct

 Tags geometry, question, time

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post ron246 Trigonometry 1 May 20th, 2014 09:07 PM frogboxer Elementary Math 1 July 8th, 2013 07:20 AM sterces Elementary Math 4 September 7th, 2010 11:53 AM Darpa.Chief Physics 1 April 16th, 2010 12:57 PM peterle1 Probability and Statistics 1 February 27th, 2010 05:02 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top