Question #43f13

1. Find location of #B#:
   Given: The midpoint of #AB (1,5)# is up 3 and over 4 from #A#. #m_(AB) = 3/4#. This means #B# is up 3 and over 4 from (1,5): #B(5, 8), " and AB = 10#

2. Find location of #D#:
   Given: #m_(DA) = -4/3 = -1/(m_(AB))#, #" so " AB " perpendicular " DA = 90^@#
   From A go down 4, over 3: #D (0, -2)#
   Note: Since we don't know the length of #DA#, the location of D can vary.

3. Find location of #C#:
   Given: #CD# and #DA# are #90^@# this means they are perpendicular. #m_(DA) = -4/3 = -1/(m_(CD)), m_(CD) = 3/4 = m_(AB)#. This means #CD |\| AB#
   Given: #CD = 10#, using Pythagorean Theorem: #10^2 = (4x)^2 +(3x)^2#
   #100 = 16x^2 + 9x^2#
   #100 = 25x^2#
   #x^2 = 4#
   #x = 2#, #" so " 4x = 8#, #3x = 6#
   This means from #D# we need to go up 8, over 6 from #D# to find #C (8, 4)#
   Note: The location of #C# depends on #D#, which can vary. This means we don't know the length of #DA# or #AB#.

4. Find the #m_(BC ) = (y_2 - y_1)/(x_2 - x_1) = (8 - 4)/(5-8) = 4/-3 = -4/3# #" so " BC " is perpendicular to " AB#

So #ABCD = "rectangle"#