Base of an isosceles triangle #XYZ# is #XZ# and coordinates of its vertices are mentioned as #X(-2,3),Y(1,5)# and #Z(square,2)#. What is the missing coordinate?

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Answer 1 · 2017-10-11T06:31:04

Missing coordinates are #Z(-1,2)# or #Z(3,2)#

As the base of the isoceles triangle is #XZ#, we would have #XY=YZ#

Let the coordinates be #X(-2,3)#, #Y(1,5)# and #Z(x,2)#

Then as #XY=YZ#, we have

#sqrt((1-(-2))^2+(5-3)^2)=sqrt((x-1)^2+(2-5)^2)#

or #3^2+2^2=(x-1)^2+(-3)^2#

or #9+4=(x-1)^2+9#

or #(x-1)^2-4=0#

or #(x-1+2)(x-1-2)=0#

i.e. #x=-1# or #3#

Hence missing coordinates are #Z(-1,2)# or #Z(3,2)#