My Math Forum Prove that these 3 points are in the same half circle.

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 October 3rd, 2019, 01:20 AM #1 Newbie   Joined: Sep 2019 From: Fier, Albania Posts: 3 Thanks: 0 Prove that these 3 points are in the same half circle. Prove that A(-10;-12) B(6;8) C(-2;-14) are in the same half circle. Please help; been torturing myself all morning. Last edited by skipjack; October 3rd, 2019 at 02:03 PM.
 October 3rd, 2019, 02:58 AM #2 Senior Member     Joined: Feb 2010 Posts: 714 Thanks: 151 The coordinates of point $\displaystyle B$ are obscured by an emoticon. If the coordinates of $\displaystyle B$ are ($\displaystyle 6,8$) then the answer is that the three points are not on the same half circle. Clearly they are on a circle. Thanks from idontknow
 October 3rd, 2019, 03:31 AM #3 Newbie   Joined: Sep 2019 From: Fier, Albania Posts: 3 Thanks: 0 How do I prove that then? That they are not in a half-circle.
 October 3rd, 2019, 03:43 AM #4 Senior Member   Joined: Mar 2015 From: Universe 2.71828i3.14159 Posts: 132 Thanks: 49 Math Focus: Area of Circle Calculate the center of the circle. Draw line from one of a given point to center. Check if two other points lie on the same side of the line.
 October 3rd, 2019, 03:44 AM #5 Senior Member   Joined: Jun 2019 From: USA Posts: 310 Thanks: 162 Plotting the points, it is too close to tell decisively. From geometry: If three vertices of a triangle are on the same semicircle, there will be one angle of 90° or more. We could use trig, but we can get away with Pythagoras. AB = (16,20), |AB| = √(16²+20²) = √656 = 25.6124 BC = (-8,-22), |BC| = √(8²+22²) = √548 = 23.4094 CA = (-8,2), |CA| = √(8²+2²) = √68 = 8.2462 If this were a right triangle, then the hypotenuse |AB| would be √(|BC|²+|CA|²) = √(548+68) = √616 = 24.8193. |AB| is longer than this, so
 October 3rd, 2019, 03:50 AM #6 Senior Member   Joined: Mar 2015 From: Universe 2.71828i3.14159 Posts: 132 Thanks: 49 Math Focus: Area of Circle The formula for a circle is $x^2+y^2+ax+by+c=0$ $x=-10, \; y=-12 \Rightarrow 100+144-10a-12b+c=0 \Rightarrow 10a+12b-c=244$ $x=6, \; y=8 \Rightarrow 36+64+6a+8b+c=0 \Rightarrow 6a+8b+c=-100$ $x=-2, \; y=-14 \Rightarrow 4+196-2a-14b+c=0 \Rightarrow 2a+14b-c=200$
 October 3rd, 2019, 03:58 AM #7 Senior Member   Joined: Jun 2019 From: USA Posts: 310 Thanks: 162 ^ That could work. Seems like a lot more effort, though.
October 3rd, 2019, 02:41 PM   #8
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Quote:
 Originally Posted by mrtwhs . . . the answer is that the three points are not on the same half circle.
That's incorrect - they do lie on the same semicircle.

The points are the vertices of a triangle. It suffices to show that the triangle has an obtuse angle (or right angle), as already shown above.

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