# 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?

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

${m}_{A} = \frac{4 - 1}{7 - 2} = \frac{3}{5}$

The direction of A axis:

${m}_{A \vdash} = - \frac{1}{m} _ A = - \frac{5}{3}$

The equation of A axis:

$y - \frac{5}{2} = - \frac{5}{3} \left(x - \frac{9}{2}\right)$

$5 x + 3 y - 30 = 0$

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:

$A = \frac{\left\mid 2 \left(4 - {y}_{0}\right) + 7 \left({y}_{0} - 1\right) + \frac{30 - 3 {y}_{0}}{5} \cdot \left(1 - 4\right) \right\mid}{2}$

so

$8 - 2 {y}_{0} + 7 {y}_{0} - 7 - 18 + \frac{9}{5} {y}_{0} = \pm 36$

$\frac{34}{5} {y}_{0} = 17 \pm 36$

${y}_{0} = \frac{5}{2} \pm \frac{90}{17}$

${y}_{1} = \frac{265}{34} \setminus \setminus \setminus \setminus , \setminus \setminus \setminus \setminus {y}_{2} = - \frac{95}{34}$

${x}_{1} = \frac{30 - 3 {y}_{1}}{5} = \frac{45}{34} \setminus \setminus \setminus \setminus , \setminus \setminus \setminus \setminus {x}_{2} = \frac{30 - 3 {y}_{2}}{5} = \frac{261}{34}$