Question

The vertices of triangle ABC are A (7, 7), B (-4, 3) and C (2, -5). Calculate the length of:

a) the longest side of triangle ABC

b) the line AM, where M is the midpoint of the side opposite A.

Answer 1

Answer:

1. AC  2. $\sqrt{73}$ units

Step-by-step explanation:

A(7,7), B(-4,3), C(2,-5) are the given points. Now, finding the distance between these points:

AB = $\sqrt{[7-(-4)^{2}] + [7-3]^{2} }$

     = $\sqrt{137}$ units

AC = $\sqrt{[7-2]^{2}+[7-(-5)]^{2} }$

     = 13 or $\sqrt{169}$ units

BC = $\sqrt{[(-4)-2]^{2}+[3-(-5)]^{2} }$

     = 10 or $\sqrt{100}$ units

Among these, the longest line is AC with value 13 or $\sqrt{169}$ units.

The line opposite to A is BC and its midpoint is M. The coordinates of midpoint are:

M(x,y) = (-4+2)/2 , {3-(-5)}/2}

          = (-1 ,4)

The length of line AM will be the distance between A(7,7) and M(-1,4):

AM = $\sqrt{[7-(-1)]^{2}+[7-4]^{2} }$

      =  $\sqrt{73}$ units

PS: Make my answer the brainliest please. Thankss