Your diagram should show that you have a right angle at B , but let's prove it anyway.
slope AB = (2-1)/(5-1) = 1/4
slope of BC = (6-2)/(4-5) = -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 + (2-3.5)^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
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....
5 answers
What is the solution when the formula used is distance between points??
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 + (3.5-1)^2 )
= √( 1.5^2 + 2.5^2)
= √2.25 + 6.25)
= √8.5 ---- see above
= appr 2.9155
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 + (3.5-1)^2 )
= √( 1.5^2 + 2.5^2)
= √2.25 + 6.25)
= √8.5 ---- see above
= appr 2.9155
I mean is to get the distance of A B C you used distance formula not slope...
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 right-angled.
Other than that, finding the slope had nothing to do with the calculations of the distances.
All I did with the slope calculation was to show that the triangle is indeed right-angled.
Other than that, finding the slope had nothing to do with the calculations of the distances.