# Two corners of an isosceles triangle are at (8 ,3 ) and (5 ,4 ). If the triangle's area is 4 , what are the lengths of the triangle's sides?

Jun 27, 2016

The length of the sides are $\sqrt{10} , \sqrt{10} , \sqrt{8}$ and the points are $\left(8 , 3\right) , \left(5 , 4\right) \mathmr{and} \left(6 , 1\right)$

#### Explanation:

Let the points of the triangle be $\left({x}_{1} , {y}_{1}\right) , \left({x}_{2} , {y}_{2}\right) , \left({x}_{3} , {y}_{3}\right) .$

Area of triangle is A = ((x_1(y_2 – y_3) + x_2(y_3 – y_1) + x_3(y_1 – y_2))/2)

Given $A = 4 , \left({x}_{1} , {y}_{1}\right) = \left(8 , 3\right) , \left({x}_{2} , {y}_{2}\right) = \left(5 , 4\right)$

Substituting we have the below Area equation:

((8(4 – y_3) + 5(y_3 – 3) + x_3(3 – 4))/2 )= 4

((8(4 – y_3) + 5(y_3 – 3) + x_3(3 – 4)) = 8

(32 – 8y_3) + (5y_3 – 15) + (-1x_3) = 8

17 – 3y_3 -x_3 = 8

– 3y_3 -x_3 = (8-17)

– 3y_3 -x_3 = -9

$3 {y}_{3} + {x}_{3} = 9$ ----> Equation 1

Distance between points $\left(8 , 3\right) , \left(5 , 4\right)$ using distance formula is

$\sqrt{{\left(8 - 5\right)}^{2} + {\left(3 - 4\right)}^{2}}$ = $\sqrt{{3}^{2} + {\left(- 1\right)}^{2}}$ = $\sqrt{10}$

Distance between points $\left({x}_{3} , {y}_{3}\right) , \left(5 , 4\right)$ using distance formula is
$\sqrt{{\left({x}_{3} - 5\right)}^{2} + {\left({y}_{3} - 4\right)}^{2}}$ = $\sqrt{10}$

Squaring both sides and subsituting ${x}_{3} = 9 - 3 {y}_{3}$ from equation 1, we get a quadratic equation.

${\left(9 - 3 {y}_{3} - 5\right)}^{2} + {\left({y}_{3} - 4\right)}^{2} = 0$

${\left(4 - 3 {y}_{3}\right)}^{2} + {\left({y}_{3} - 4\right)}^{2} = 0$

Factorizing this, we get $\left(y - 1\right) \left(10 y - 22\right) = 0$

y = 1 or y = 2.2 . y=2.2 can be discarded. Hence, the third point has to be (6,1).

By calculating the distances for points $\left(8 , 3\right) , \left(5 , 4\right) \mathmr{and} \left(6 , 1\right)$, we get $\sqrt{8}$ for the length of the base.