3 points have coordinates A(-2,1), B(10,6), C(a,-6). given that AB=BC, find the possible values of a. if the figure ABCD is a rhombus, the find coordinates of D.
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distance from A to B is
d1 = sqrt((10+2)^2+ (6-1)^2) = sqrt(169) = 13
Distance from B to C is
d2 = sqrt((a-10)^2 + (-6-6)^2 ) = sqrt(a^2 -20a+ 244)
d1=d2 then
sqrt(a^2 -20a+ 244) = 13
a^2 -20a+ 75=0
(a-5)(a-15) = 0
then a = 5 or a = 15
d1 = sqrt((10+2)^2+ (6-1)^2) = sqrt(169) = 13
Distance from B to C is
d2 = sqrt((a-10)^2 + (-6-6)^2 ) = sqrt(a^2 -20a+ 244)
d1=d2 then
sqrt(a^2 -20a+ 244) = 13
a^2 -20a+ 75=0
(a-5)(a-15) = 0
then a = 5 or a = 15