An isosceles triangle has sides A, B, and C with sides B and C being equal in length. If side A goes from #(7 ,4 )# to #(2 ,1 )# and the triangle's area is #15 #, what are the possible coordinates of the triangle's third corner?

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Answer 1 · 2017-07-22T08:31:09

The coordinates of the third corner are #(7.14,-1.9)# or #(1.85,6.92)#

The length of the base of the isoceles triangle is

#b=sqrt((7-2)^2+(4-1)^2)=sqrt(25+9)=sqrt34#

The area of the triangle is

#area=1/2bh=15#

The altitude is #=30/b=30/sqrt34#

The mid-point of the base is

#=((7+2)/2,(4+1)/2)=(9/2,5/2)#

The slope of the base is #m=(1-4)/(2-7)=-3/-5=3/5#

The slope of the altitude is #m'=-1/m=-5/3#

The equation of the altitude is

#y-5/2=-5/3(x-9/2)#

#y=-5/3x+15/2+5/2=-5/3x+10#........................#(1)#

Let the coordinates of the third corner be #=(x,y)#

Then,

#(x-9/2)^2+(y-5/2)^2=(30/sqrt34)^2#

#(x-9/2)^2+(y-5/2)^2=900/34=26.5#..................#(2)#

Solving for #x# and #y# in equations #(1)# and #(2)#

#(x-4.5)^2+(-5/3x+10-2.5)^2=26.5#

#(x-4.5)^2+(-5/3x+7.5)^2=26.5#

#x^2-9x+20.25+2.78x^2-25x+56.25-26.5=0#

#3.78x^2-34x+50=0#

Solving this quadratic equation

#x=(34+-sqrt(34^2-4*3.78*50))/(2*3.78)#

#x=(34+-sqrt400)/(7.56)=(34+-20)/7.56#

#x_1=7.14# or #x_2=1.85#

#y_1=-5/3*7.14+10=-1.9#

#y_2=-5/3*1.85+10=6.92#