# An isosceles triangle has sides A, B, and C with sides B and C being equal in length. If side A goes from (7 ,1 ) to (2 ,9 ) and the triangle's area is 18 , what are the possible coordinates of the triangle's third corner?

##### 1 Answer

There are two possible points at $\textcolor{b l u e}{\left(\frac{1377}{178} \text{,} \frac{625}{89}\right)}$ and
$\textcolor{red}{\left(\frac{225}{178} \text{,} \frac{265}{89}\right)}$

#### Explanation:

The solution makes use of the formula for the area of a triangle

Area $= \frac{1}{2} \left[\begin{matrix}{x}_{1} & {x}_{2} & {x}_{3} & {x}_{1} \\ {y}_{1} & {y}_{2} & {y}_{3} & {y}_{1}\end{matrix}\right]$

positive area for counterclockwise arrangement of points
negative area for clockwise arrangements of points

Let (x_1, y_1) be the unknown point and Let ${P}_{2} \left(2 , 9\right)$ and ${P}_{3} \left(7 , 1\right)$ and P_1(x_1, y_1) and Area$= 18$

Area $= \frac{1}{2} \left[\begin{matrix}{x}_{1} & {x}_{2} & {x}_{3} & {x}_{1} \\ {y}_{1} & {y}_{2} & {y}_{3} & {y}_{1}\end{matrix}\right]$

$+ 18 = \frac{1}{2} \left[\begin{matrix}{x}_{1} & 2 & 7 & {x}_{1} \\ {y}_{1} & 9 & 1 & {y}_{1}\end{matrix}\right]$

$18 \left(2\right) = \left(9 {x}_{1} + 2 + 7 {y}_{1} - 2 {y}_{1} - 63 - {x}_{1}\right)$

$36 = 8 {x}_{1} + 5 {y}_{1} - 61$

$8 {x}_{1} + 5 {y}_{1} = 97 \text{ }$first equation

For the equal sides because the triangle is isosceles, we have

${\left({x}_{1} - 2\right)}^{2} + {\left({y}_{1} - 9\right)}^{2} = {\left({x}_{1} - 7\right)}^{2} + {\left({y}_{1} - 1\right)}^{2}$
${x}_{1}^{2} - 4 {x}_{1} + 4 + {y}_{1}^{2} - 18 {y}_{1} + 81 = {x}_{1}^{2} - 14 {x}_{1} + 49 + {y}_{1}^{2} - 2 {y}_{1} + 1$
Simplify
$- 4 {x}_{1} + 4 - 18 {y}_{1} + 81 = - 14 {x}_{1} + 49 - 2 {y}_{1} + 1$
$10 {x}_{1} - 16 {y}_{1} + 85 - 50 = 0$
$10 {x}_{1} - 16 {y}_{1} + 35 = 0 \text{ }$second equation

Simultaneous solution using first and second equations results to
$\textcolor{red}{\left({x}_{1} , {y}_{1}\right) = \left(\frac{1377}{178} , \frac{625}{89}\right)}$

There is another point which may be solved by using the negative area

Area $= \frac{1}{2} \left[\begin{matrix}{x}_{1} & {x}_{2} & {x}_{3} & {x}_{1} \\ {y}_{1} & {y}_{2} & {y}_{3} & {y}_{1}\end{matrix}\right]$
$- 18 = \frac{1}{2} \left[\begin{matrix}{x}_{1} & 2 & 7 & {x}_{1} \\ {y}_{1} & 9 & 1 & {y}_{1}\end{matrix}\right]$

$- 18 \left(2\right) = \left(9 {x}_{1} + 2 + 7 {y}_{1} - 2 {y}_{1} - 63 - {x}_{1}\right)$

$- 36 = 8 {x}_{1} + 5 {y}_{1} - 61$

$8 {x}_{1} + 5 {y}_{1} = 25 \text{ }$third equation

Simultaneous solution using second and third equations results to
$\textcolor{red}{\left({x}_{1} , {y}_{1}\right) = \left(\frac{225}{178} , \frac{265}{89}\right)}$

God bless....I hope the explanation is useful.