Question #5aed5

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Question #5aed5

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Answer 1 · 2016-10-07T00:05:55

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

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

$dist ((x_{1}, y_{1}), (x_{2}, y_{2})) = \sqrt{{(x_{2} - x_{1})}_{2}^{2} + {(y_{2} - y_{1})}_{2}^{2}}$ $"dist"((x_1, y_1)"",(x_2, y_2)) = sqrt((x_2-x_1)^2+(y_2-y_1)^2)$

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

$A B^{2} = {(3 - 2)}^{2} + {(5 - 0)}^{2} = 26$ $AB^2 = (3-2)^2+(5-0)^2 = 26$
$A C^{2} = {(0 - 2)}^{2} + {(y - 0)}^{2} = y^{2} + 4$ $AC^2 = (0-2)^2+(y-0)^2 = y^2+4$
$B C^{2} = {(0 - 3)}^{2} + {(y - 5)}^{2} = y^{2} - 10 y + 34$ $BC^2 = (0-3)^2+(y-5)^2 = y^2-10y+34$

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

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

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

$\Rightarrow 10 y = 4$ $=> 10y = 4$

$\Rightarrow y = \frac{2}{5}$ $=> y = 2/5$

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

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

$\Rightarrow 10 y = 56$ $=> 10y = 56$

$\Rightarrow y = \frac{28}{5}$ $=> y = 28/5$

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

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

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

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

$\Rightarrow (y - 2) (y - 3) = 0$ $=> (y-2)(y-3) = 0$

$\Rightarrow y - 2 = 0 or y - 3 = 0$ $=> y-2 = 0 or y-3 = 0$

$\Rightarrow y = 2 or y = 3$ $=> y = 2 or y = 3$

By the Pythagorean theorem's converse, we know that all of the above $y$ $y$ 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 $y$ $y$ which work. Thus we have the solution set

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

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