An isosceles triangle has sides A, B, and C with sides B and C being equal in length. If side A goes from (7 ,5 ) to (8 ,2 ) and the triangle's area is 27 , what are the possible coordinates of the triangle's third corner?

1 Answer
Jul 1, 2017

The coordinates are (23.7,8.9) and (-8.7,-1.9)

Explanation:

The length of side A=sqrt((7-8)^2+(5-2)^2)=sqrt10

Let the height of the triangle be =h

The area of the triangle is

1/2*sqrt10*h=27

The altitude of the triangle is h=(27*2)/sqrt10=54/sqrt10

The mid-point of A is (15/2,7/2)

The gradient of A is =(2-5)/(8-7)=-3

The gradient of the altitude is =1/3

The equation of the altitude is

y-7/2=1/3(x-15/2)

y=1/3x-5/2+7/2=1/3x+1

The circle with equation

(x-15/2)^2+(y-7/2)^2=54^2/10=291.6

The intersection of this circle with the altitude will give the third corner.

(x-15/2)^2+(1/3x+1-7/2)^2=291.6

x^2-15x+225/4+1/9x^2-5/3x+25/4=291.6

1.11x^2-16.7x-229.1=0

We solve this quadratic equation

x=(16.7+-sqrt(16.7^2+4*1.11*229.1))/(2*1.11)

=(16.7+-36)/2.22

x_1=23.74

x_2=-8.69

The points are (23.7,9.65) and (-8.7,-1.9)

graph{(y-1/3x-1)((x-7.5)^2+(y-3.5)^2-291.6)((x-7)^2+(y-5)^2-0.05)((x-8)^2+(y-2)^2-0.05)(y-5+3(x-7))=0 [-12, 28, -10, 10]}