Question

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?

Answer 1

`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`