# Question 5aed5

Oct 7, 2016

$y \in \left\{\frac{2}{5} , 2 , 3 , \frac{28}{5}\right\}$

#### Explanation:

The Pythagorean theorem's converse is also true, and states that if the sum of the squares of two sides of a triangle equals the square of the third side, then the triangle is a right triangle.

Using the distance formula

"dist"((x_1, y_1)"",(x_2, y_2)) = sqrt((x_2-x_1)^2+(y_2-y_1)^2)#

we can calculate the squares of the lengths of each side:

$A {B}^{2} = {\left(3 - 2\right)}^{2} + {\left(5 - 0\right)}^{2} = 26$
$A {C}^{2} = {\left(0 - 2\right)}^{2} + {\left(y - 0\right)}^{2} = {y}^{2} + 4$
$B {C}^{2} = {\left(0 - 3\right)}^{2} + {\left(y - 5\right)}^{2} = {y}^{2} - 10 y + 34$

Now we can consider all cases in which two of those sum to the third, and what values of $y$ make that true.

Case 1: $A {B}^{2} + A {C}^{2} = B {C}^{2}$

$\implies {y}^{2} + 30 = {y}^{2} - 10 y + 34$

$\implies 10 y = 4$

$\implies y = \frac{2}{5}$

Case 2: $A {B}^{2} + B {C}^{2} = A {C}^{2}$

$\implies {y}^{2} - 10 y + 60 = {y}^{2} + 4$

$\implies 10 y = 56$

$\implies y = \frac{28}{5}$

Case 3: $A {C}^{2} + B {C}^{2} = A {B}^{2}$

$\implies 2 {y}^{2} - 10 y + 38 = 26$

$\implies 2 {y}^{2} - 10 y + 12 = 0$

$\implies {y}^{2} - 5 y + 6 = 0$

$\implies \left(y - 2\right) \left(y - 3\right) = 0$

$\implies y - 2 = 0 \mathmr{and} y - 3 = 0$

$\implies y = 2 \mathmr{and} y = 3$

By the Pythagorean theorem's converse, we know that all of the above $y$ values result in right triangles, and as those are the only values which result in the Pythagorean theorem holding true, we know that there are no other values for $y$ which work. Thus we have the solution set

$y \in \left\{\frac{2}{5} , 2 , 3 , \frac{28}{5}\right\}$