# An isosceles triangle has sides A, B, and C with sides B and C being equal in length. If side A goes from (2 ,9 ) to (8 ,5 ) and the triangle's area is 48 , what are the possible coordinates of the triangle's third corner?

Jun 30, 2018

$\left(\frac{161}{13} , \frac{235}{13}\right) , \mathmr{and} , \left(\frac{31}{13} , - \frac{53}{13}\right)$.

#### Explanation:

Let the third corner be at $\left(x , y\right)$.

Thus, the vertices of the $\Delta$ are, $\left(x , y\right) , \left(2 , 9\right) \mathmr{and} \left(8 , 5\right)$.

Hence, the area of the $\Delta$ is given by, $\frac{1}{2} | D | , \text{ where, }$

$D = | \left(x , y , 1\right) , \left(2 , 9 , 1\right) , \left(8 , 5 , 1\right) |$,

$= x \left(9 - 5\right) - y \left(2 - 8\right) + 1 \left(10 - 72\right)$,

$= 4 x + 6 y - 62$,

$= 2 \left(2 x + 3 y - 31\right)$.

But, this area is $48. \therefore | \left(2 x + 3 y - 31\right) | = 48$.

$\therefore 2 x + 3 y - 31 = \pm 48 , i . e . ,$

$2 x + 3 y = 79. \ldots \ldots \ldots . . \left(1\right) , \mathmr{and} , 2 x + 3 y = - 17. \ldots \ldots \ldots . . \left(2\right)$.

Also, the length of side $B = \sqrt{{\left(x - 2\right)}^{2} + {\left(y - 9\right)}^{2}}$,

&, that of $C = \sqrt{{\left(x - 8\right)}^{2} + {\left(y - 5\right)}^{2}}$.

Then, $B = C \ldots \ldots \ldots \ldots \ldots \ldots \text{[Given]}$,

$\Rightarrow {x}^{2} + {y}^{2} - 4 x - 18 y + 85 = {x}^{2} + {y}^{2} - 16 x - 10 y + 89$.

$\Rightarrow 12 x - 8 y = 4 , \mathmr{and} , 3 x - 2 y = 1. \ldots \ldots \ldots \ldots \ldots \ldots . \left(3\right)$.

Solving (1), &, (3)," gives, "(x,y)=(161/13,235/13)," whereas, "

 (2), &, (3), (x,y)=(-31/13,-53/13).

Thus, the third corner is $\left(\frac{161}{13} , \frac{235}{13}\right) , \mathmr{and} , \left(\frac{31}{13} , - \frac{53}{13}\right)$.

Enjoy Maths.!