# An isosceles triangle has sides A, B, and C with sides B and C being equal in length. If side A goes from (4 ,1 ) to (8 ,5 ) and the triangle's area is 64 , what are the possible coordinates of the triangle's third corner?

$\left(22 , - 13\right)$ or $\left(- 10 , 19\right)$

#### Explanation:

Given in isosceles $\setminus \triangle A B C$, the side $A$ opposite to the vertex $A$ goes through the vertices $B \left(4 , 1\right)$ & $C \left(8 , 5\right)$.

$B C = \setminus \sqrt{{\left(4 - 8\right)}^{2} + {\left(1 - 5\right)}^{2}}$

$= 4 \setminus \sqrt{2}$

Mid-point say $M$ of side $B C : \setminus \setminus \setminus M \left(\setminus \frac{4 + 8}{2} , \setminus \frac{1 + 5}{2}\right) \setminus \equiv \left(6 , 3\right)$

Let $h$ be the length of altitude drawn from vertex $A$ to the side $B C$ then area of $\setminus \triangle A B C$

$\frac{1}{2} \left(B C\right) h = 64$

$\frac{1}{2} \left(4 \setminus \sqrt{2}\right) h = 64$

$h = 16 \setminus \sqrt{2}$

Let the third vertex be $A \left(x , y\right)$ then the altitude $A M$ drawn from vertex $A \left(x , y\right)$ to the midpoint $M \left(6 , 3\right)$ of side $B C$ will be perpendicular to the side $B C$ hence the condition of perpendicularity of two lines

$\setminus \frac{y - 3}{x - 6} \setminus \times \setminus \frac{5 - 1}{8 - 4} = - 1$

$y = - x + 9 \setminus \setminus \ldots \ldots \ldots . . \left(1\right)$

Now, the distance between the vertex $A \left(x , y\right)$ & mid-points $M \left(6 , 3\right)$ of side BC must be equal to the length of altitude $h = 16 \setminus \sqrt{2}$ drawn from vertex A to the side BC. Now, using distance formula

$\setminus \sqrt{{\left(x - 6\right)}^{2} + {\left(y - 3\right)}^{2}} = 16 \setminus \sqrt{2}$

${\left(x - 6\right)}^{2} + {\left(y - 3\right)}^{2} = 512$

${\left(x - 6\right)}^{2} + {\left(- x + 9 - 3\right)}^{2} = 512 \setminus \setminus \quad \left(\setminus \because y = - x + 9 \setminus \setminus \textrm{\mathfrak{o} m \left(1\right)}\right)$

${\left(x - 6\right)}^{2} = 256$

$x - 6 = \setminus \pm 16$

$x = 6 \setminus \pm 16$

$x = 22 , - 10$

setting these values of $x$ into (1), we get corresponding values of $y$ as follows

$y = - 22 + 9 = - 13$ &

$y = - \left(- 10\right) + 9 = 19$

Hence the coordinates of third vertex are $A \left(22 , - 13\right)$ or $A \left(- 10 , 19\right)$