Posted by Tsunayoshi on Monday, June 9, 2014 at 8:27am.
Find the coordinates of the midpoint of the hypotenuse of the right triangle whose vertices are A (1,1) B (5,2) C (4,6) and show that it is equidistant of each of the vertices.....
And can you please tell me how to prove it in drawing....

Algebra  Reiny, Monday, June 9, 2014 at 9:10am
Your diagram should show that you have a right angle at B , but let's prove it anyway.
slope AB = (21)/(51) = 1/4
slope of BC = (62)/(45) = 4
YES, AB is perpendicular to BC, thus AC is a hypotenuse
midpoint of AC = M( (1+4)/2 , (1+6)/2) = M(5/2 , 7/2)
= M(2.5 , 3/5)
AM = √(1.5^2 + 2.5^2 = √8.5
CM = √(1.5^2 + 2.5^2) = √8.5 , clearly has to be, since we found the midpoint
BM = √((5  2.5)^2 + (23.5)^2 ) = √8.5
YUP, all is good
If you make a good sketch, you should be able to show it by counting the displacements along the grid pattern 
Algebra  tsunayoshi, Monday, June 9, 2014 at 9:31am
What is the solution when the formula used is distance between points??

Algebra  Reiny, Monday, June 9, 2014 at 10:04am
I don't understand your question.
I DID use the distance between two points formula, I just did not show the subtraction since that is easily done in your head.
here is one of them with all steps :
AM = √( 2.5  1)^2 + (3.51)^2 )
= √( 1.5^2 + 2.5^2)
= √2.25 + 6.25)
= √8.5  see above
= appr 2.9155 
Algebra  Tsunayoshi, Monday, June 9, 2014 at 10:25am
I mean is to get the distance of A B C you used distance formula not slope...

Algebra  Reiny, Monday, June 9, 2014 at 11:06am
I think the main purpose of the question, even though they did not say that, was to show that the midpoint of the hypotenuse is equidistant from the 3 vertices.
All I did with the slope calculation was to show that the triangle is indeed rightangled.
Other than that, finding the slope had nothing to do with the calculations of the distances.