# 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 ,5 ) and the triangle's area is 32 , what are the possible coordinates of the triangle's third corner?

Possible coordinates are$\left(\frac{881}{82} , \frac{443}{41}\right)$ and $\left(- \frac{143}{82} , - \frac{197}{41}\right)$

#### Explanation:

We use the formula for the area A

$A = \frac{1}{2} \left[\begin{matrix}{x}_{a} & {x}_{b} & {x}_{c} & {x}_{a} \\ {y}_{a} & {y}_{b} & {y}_{c} & {y}_{a}\end{matrix}\right]$

$A = \frac{1}{2} \left[{x}_{a} \cdot {y}_{b} + {x}_{b} \cdot {y}_{c} + {x}_{c} \cdot {y}_{a} - {x}_{b} \cdot {y}_{a} - {x}_{c} \cdot {y}_{b} - {x}_{a} \cdot {y}_{c}\right]$

We have to consider two points on each side of side $a$

For the first point $A$

$A = \frac{1}{2} \left[\begin{matrix}{x}_{a} & {x}_{b} & {x}_{c} & {x}_{a} \\ {y}_{a} & {y}_{b} & {y}_{c} & {y}_{a}\end{matrix}\right]$

$A = \frac{1}{2} \left[{x}_{a} \cdot {y}_{b} + {x}_{b} \cdot {y}_{c} + {x}_{c} \cdot {y}_{a} - {x}_{b} \cdot {y}_{a} - {x}_{c} \cdot {y}_{b} - {x}_{a} \cdot {y}_{c}\right]$

$32 = \frac{1}{2} \left[\begin{matrix}{x}_{a} & 2 & 7 & {x}_{a} \\ {y}_{a} & 5 & 1 & {y}_{a}\end{matrix}\right]$

$32 = \frac{1}{2} \left[5 {x}_{a} + 2 + 7 {y}_{a} - 2 {y}_{a} - 35 - {x}_{a}\right]$

$64 - 2 + 35 = 4 {x}_{a} + 5 {y}_{a}$

$97 = 4 {x}_{a} + 5 {y}_{a}$

This is an isosceles triangle so that the third vertex $A \left({x}_{a} , {y}_{a}\right)$ passes through the perpendicular bisector of side $a$

Perpendicular bisector equation is passing through the midpoint of side $a$ which $M \left(\frac{9}{2} , 3\right)$ with slope ${m}_{p} = \frac{5}{4}$

$y - 3 = \frac{5}{4} \left(x - \frac{9}{2}\right)$

$10 x - 8 y = 21$

Also this is

$10 {x}_{a} - 8 {y}_{a} = 21$

Solving Vertex $A$ with equation $97 = 4 {x}_{a} + 5 {y}_{a}$

$4 {x}_{a} + 5 {y}_{a} = 97$
$10 {x}_{a} - 8 {y}_{a} = 21$

Simultaneous solution results to

$\textcolor{red}{A \left({x}_{a} , {y}_{a}\right) = \left(\frac{881}{82} , \frac{443}{41}\right)}$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

For the second point $A$ on the other side of side $a$

$A = \frac{1}{2} \left[\begin{matrix}{x}_{a} & {x}_{c} & {x}_{b} & {x}_{a} \\ {y}_{a} & {y}_{c} & {y}_{b} & {y}_{a}\end{matrix}\right]$

$A = \frac{1}{2} \left[{x}_{a} \cdot {y}_{b} + {x}_{b} \cdot {y}_{c} + {x}_{c} \cdot {y}_{a} - {x}_{b} \cdot {y}_{a} - {x}_{c} \cdot {y}_{b} - {x}_{a} \cdot {y}_{c}\right]$

$32 = \frac{1}{2} \left[\begin{matrix}{x}_{a} & 7 & 2 & {x}_{a} \\ {y}_{a} & 1 & 5 & {y}_{a}\end{matrix}\right]$

$32 = \frac{1}{2} \left[5 {x}_{a} + 2 + 7 {y}_{a} - 2 {y}_{a} - 35 - {x}_{a}\right]$

$64 + 2 - 35 = - 4 {x}_{a} - 5 {y}_{a}$

$4 {x}_{a} + 5 {y}_{a} = - 31$

This is an isosceles triangle so that the other third vertex $A \left({x}_{a} , {y}_{a}\right)$ passes through the perpendicular bisector of side $a$

Perpendicular bisector equation is passing through the midpoint of side $a$ which $M \left(\frac{9}{2} , 3\right)$ with slope ${m}_{p} = \frac{5}{4}$

$y - 3 = \frac{5}{4} \left(x - \frac{9}{2}\right)$

$10 x - 8 y = 21$

Also this is

$10 {x}_{a} - 8 {y}_{a} = 21$

Solving the other Vertex $A$ with equation $4 {x}_{a} + 5 {y}_{a} = - 31$

$4 {x}_{a} + 5 {y}_{a} = - 31$
$10 {x}_{a} - 8 {y}_{a} = 21$

Simultaneous solution results to

$\textcolor{red}{A \left({x}_{a} , {y}_{a}\right) = \left(- \frac{143}{82} , - \frac{197}{41}\right)}$

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