An isosceles triangle has sides A, B, and C with sides B and C being equal in length. If side A goes from #(7 ,1 )# to #(8 ,5 )# and the triangle's area is #27 #, what are the possible coordinates of the triangle's third corner?

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Answer 1 · 2016-12-15T18:13:31

#(6.55882, 3.23529) and (8.44118, 2.76471)#

Let's begin by finding the length of side "a":

#a = sqrt((8-7)^2 + (5 - 1)^2)#

#a = sqrt(1^2 + 4^2)#

#a = sqrt(17)#

Use side "a" as the base of the triangle, find the height:

#Area = 1/2base xx height#

#2 = 1/2sqrt(17) xx height#

#height = (4sqrt(17))/17#

The height must bisect side "a" at the point #(7.5, 3)#

We can used the distance formula to write one equation for the two possible points:

#(4sqrt(17))/17 = sqrt((x - 7.5)^2 + (y - 3)^2)#

Square both sides:

#16/17 = (x - 7.5)^2 + (y - 3)^2" [1]"#

We need one more equation. Find the slope of side "a":

#m = (5 - 1)/(8 - 7) = 4#

Because the height is perpendicular to side "a", its slope is: #-1/4#

We can use the slope and the point to find the equation of line for the height:

#3 = -1/4(7.5) + b#

#b = 39/8#

The equation for the height is:

#y = -1/4x + 39/8" [2]"#

Solving equations [1] and [2] is very long and tedious process. I gave them to WolframAlpha. Here are the points:

#(6.55882, 3.23529) and (8.44118, 2.76471)#