An isosceles triangle has sides A, B, and C with sides B and C being equal in length. If side A goes from #(4 ,9 )# to #(8 ,5 )# and the triangle's area is #48 #, what are the possible coordinates of the triangle's third corner?

Question

1166 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-04T14:00:38

Coordinates of third corner A of the triangle is #(16/5, 21/5)#

enter image source here

Area = 48

#a = sqrt ( (4-8)^2 + (9-5)^2) = sqrt 32 = 4 sqrt 2#

#Area = (1/2) a * h#

#48 = (1/2) * 4 sqrt 2 * h#

#h = 48/ (2 sqrt 2) = 12 sqrt 2#

#tan theta = h / (a/2) = (12 sqrt 2) / (2 sqrt 2) = 6#

Equation of line passing through point B and having slope 6 is
#y - 9 = 6 * (x - 4)#
#6x - y = 15#. Eqn (1)

Slope of #BC = (5-9) / (8-4) = -1#
Slope of altitude h passing through mid point of BC is 1

Mid point of BC is #((8+4)/2, (9+5)/2) = (6, 7)#

Equation of altitude h is
#(y - 7) = 1 * (x - 6)#
#x - y = -1# Eqn (2)

Solving Eqns (1) & (2), we get coordinates of point A.

#x = 16/5, y = 21/5)#

Coordinates of the third corner A is #(16/5, 21/5)#