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 18 , what are the possible coordinates of the triangle's third corner?

1 Answer
Nov 8, 2016

P_1=(45/34;265/34)\ \ \ \ P_2=(261/34;-95/34)

Explanation:

Calculate the middle point of side A

M_A=((7+2)/2;(4+1)/2)=(9/2;5/2)

Get the direction of A

m_A=(4-1)/(7-2)=3/5

The direction of A axis:

m_(A|--)=-1/m_A=-5/3

The equation of A axis:

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

5x+3y-30=0

graph{(5x+3y-30)(3x-5y-1)((x-2)^2+(y-1)^2-0.01)((x-7)^2+(y-4)^2-0.01)((x-45/34)^2+(y-265/34)^2-0.01)((x-261/34)^2+(y+95/34)^2-0.01)=0 [-3,23,-4,9]}

Let P_0=((30-3y_0)/5;y_0) be a generic point on A axis.

The triangle area can be calculated:

A=abs(2(4-y_0)+7(y_0-1)+(30-3y_0)/5*(1-4))/2

so

8-2y_0+7y_0-7-18+9/5y_0 = +-36

34/5y_0=17+-36

y_0=5/2+-90/17

y_1=265/34\ \ \ \ ,\ \ \ \ y_2=-95/34

x_1=(30-3y_1)/5=45/34\ \ \ \ , \ \ \ \ x_2=(30-3y_2)/5=261/34