An isosceles triangle has sides A, B, and C with sides B and C being equal in length. If side A goes from #(4 ,3 )# to #(8 ,9 )# and the triangle's area is #64 #, what are the possible coordinates of the triangle's third corner?

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Answer 1 · 2016-10-19T21:22:05

#(-114/13, 206/13)# and #(270/13, -50/15)#

Side a is the base of the triangle its length is:

#a = sqrt((8 - 4)^2 + (9 - 3)^2)#

#a = sqrt(52)#

Using this and the area of the triangle when can find the altitude, h:

#64 = (1/2)sqrt(52)h

#h = 128sqrt(52)/52#

The altitude intersects the base at the midpoint of the line segment:

#((4 + 8)/2, (3 + 9)/2) = (6,6)#

The slope, m, of the base is:

#m = (9 - 3)/(8 - 4) = 6/4 = 3/2#

The slope, n, of the altitude is:

#n = -1/m = -1/(3/2) = -2/3#

The equation of the altitude is:

#y - 6 = -2/3(x - 6)#

#y = 10 - 2/3x#

Using the distance formula:

#128sqrt(52)/52 = sqrt((x - 6)^2 + (y - 6)^2)#

The two points are the intersection of the distance and the line. I used WolframAlpha to solve for the two points:

#(-114/13, 206/13)# and #(270/13, -50/15)#