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In this exercise we use the Distance Formula and the Midpoint Formula. The point M in the figure is the midpoint of the line segment AB. y . B(0, b) M C(0, 0) Alа, 0) Show that M is equidistant from the vertices of triangle ABC. The coordinates of the point M are (x, y) = We must find now the distances between point M and points A, B, and C. d(A, M) = d(в, м) 3 d(C, M) = Therefore, the following conclusion can be reached. O d(A, M) = d(B, M) + d(C, M), so M is equidistant from the vertices. O d(A, M) = d(C, M) # d(B, M), so M is equidistant from the vertices. O d(B, M) = d(C, M) # d(A, M), so M is equidistant from the vertices. The distances are all the same, so M is equidistant from the vertices. O The distances are all different, so M is equidistant from the vertices.

In this exercise we use the Distance Formula and the Midpoint Formula. The point M in the figure is the midpoint of the line segment AB. y . B(0, b) M C(0, 0) Alа, 0) Show that M is equidistant from the vertices of triangle ABC. The coordinates of the point M are (x, y) = We must find now the distances between point M and points A, B, and C. d(A, M) = d(в, м) 3 d(C, M) = Therefore, the following conclusion can be reached. O d(A, M) = d(B, M) + d(C, M), so M is equidistant from the vertices. O d(A, M) = d(C, M) # d(B, M), so M is equidistant from the vertices. O d(B, M) = d(C, M) # d(A, M), so M is equidistant from the vertices. The distances are all the same, so M is equidistant from the vertices. O The distances are all different, so M is equidistant from the vertices.

Elementary Geometry for College Students
Elementary Geometry for College Students
6th Edition
ISBN:9781285195698
Author:Daniel C. Alexander, Geralyn M. Koeberlein
Publisher:Daniel C. Alexander, Geralyn M. Koeberlein
Chapter10: Analytic Geometry
Section10.CT: Test
Problem 22CT
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In this exercise we use the Distance Formula and the Midpoint Formula.

The point M in the figure is the midpoint of the line segment AB.

 

Show that M is equidistant from the vertices of triangle ABC.

The coordinates of the point M are 
(x, y) = 
 
 
 
 
 
 
 
.


We must find now the distances between point M and points A, B, and C.
d(A, M)  = 
 
 
 
d(B, M)  = 
 
 
 
d(C, M)  = 
 
 
 


Therefore, the following conclusion can be reached.
d(A, M) = d(B, M) ≠ d(C, M), so M is equidistant from the vertices.
d(A, M) = d(C, M) ≠ d(B, M), so M is equidistant from the vertices.
    
d(B, M) = d(C, M) ≠ d(A, M), so M is equidistant from the vertices.
The distances are all the same, so M is equidistant from the vertices.
The distances are all different, so M is equidistant from the vertices.
In this exercise we use the Distance Formula and the Midpoint Formula. The point M in the figure is the midpoint of the line segment AB. y B(0, b) M C(0, 0) Alа, 0) Show that M is equidistant from the vertices of triangle ABC. The coordinates of the point M are (x, y) = We must find now the distances between point M and points A, B, and C. d(A, M) = d(в, м) d(C, M) = Therefore, the following conclusion can be reached. O d(A, M) = d(B, M) + d(C, M), so M is equidistant from the vertices. O d(A, M) = d(C, M) # d(B, M), so M is equidistant from the vertices. O d(B, M) = d(C, M) + d(A, M), so M is equidistant from the vertices. The distances are all the same, so M is equidistant from the vertices. The distances are all different, so M is equidistant from the vertices.
Transcribed Image Text:In this exercise we use the Distance Formula and the Midpoint Formula. The point M in the figure is the midpoint of the line segment AB. y B(0, b) M C(0, 0) Alа, 0) Show that M is equidistant from the vertices of triangle ABC. The coordinates of the point M are (x, y) = We must find now the distances between point M and points A, B, and C. d(A, M) = d(в, м) d(C, M) = Therefore, the following conclusion can be reached. O d(A, M) = d(B, M) + d(C, M), so M is equidistant from the vertices. O d(A, M) = d(C, M) # d(B, M), so M is equidistant from the vertices. O d(B, M) = d(C, M) + d(A, M), so M is equidistant from the vertices. The distances are all the same, so M is equidistant from the vertices. The distances are all different, so M is equidistant from the vertices.
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