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

Jan 17, 2018

Coordinates of the third corner $\textcolor{p u r p \le}{A = \left(\frac{1452}{211}\right) , \left(\frac{695}{211}\right)}$

#### Explanation: Refer diagram.

$a = \sqrt{{\left(8 - 7\right)}^{2} + {\left(2 - 5\right)}^{2}} = 3.1623$

Midpoint of base BC = $\frac{8 + 7}{2} , \frac{2 + 5}{2} = \left(\frac{15}{2} , \frac{7}{2}\right)$

Slope of BC ${m}_{B C} = \frac{2 - 5}{8 - 7} = - 3$

Slope of altitude AD of the triangle passing through the midpoint D is ${m}_{A D} = - \frac{1}{{m}_{B C}} = - \left(\frac{1}{-} 3\right) = \frac{1}{3}$

$y - \left(\frac{7}{2}\right) = \left(\frac{1}{3}\right) \cdot \left(x - \left(\frac{15}{2}\right)\right)$

$6 y - 21 = 2 x - 15$

$3 y - x = 3$. $\textcolor{b l u e}{E q n} \textcolor{w h i t e}{\times} \textcolor{b l u e}{1}$

Area of triangle $A = \left(\frac{1}{2}\right) a h$

$h = \frac{2 \cdot 36}{3.1623} = 22.7682$

Slope of AB = $\tan \theta = {m}_{A B} = \frac{h}{\frac{a}{2}} = \frac{22.7682}{\frac{3.1623}{2}} = 14.4$

Equation of AB is

$y - 5 = 14.4 \cdot \left(x - 7\right)$

$y - 14.4 x = - 95.8$. $\textcolor{b l u e}{E q n} \textcolor{w h i t e}{\times} \textcolor{b l u e}{2}$

Solving Eqns (1) & (2) we get the coordinates of vertex ‘A’

$\textcolor{p u r p \le}{A \left(\frac{1452}{211}\right) , \left(\frac{695}{211}\right)}$

Verification :

Length of AB = b = sqrt((7 - (1452/211))^2 + (5 - (695/211)^2) = 1.7103

Length of AC = b = sqrt((8-(1452/211))^2 + (2 - (695/211)’^2) = 1.7103

Value of side $A B = A C = 1.7103$.

Hence its an isosceles triangle with sides $\textcolor{g r e e n}{3.1623 , 1.7103 , 1.7103}$