Question #43f13

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Answer 1 · 2017-03-26T19:24:11

Rectangle

Find location of $B$ :
Given: The midpoint of $AB (1,5)$ is up 3 and over 4 from $A$ . $m_(AB) = 3/4$ . This means $B$ is up 3 and over 4 from (1,5): $B(5, 8), " and AB = 10$
Find location of $D$ :
Given: $m_(DA) = -4/3 = -1/(m_(AB))$ , $" so " AB " perpendicular " DA = 90^@$
From A go down 4, over 3: $D (0, -2)$
Note: Since we don't know the length of $DA$ , the location of D can vary.
Find location of $C$ :
Given: $CD$ and $DA$ are $90^@$ this means they are perpendicular. $m_(DA) = -4/3 = -1/(m_(CD)), m_(CD) = 3/4 = m_(AB)$ . This means $CD |\| AB$
Given: $CD = 10$ , using Pythagorean Theorem: $10^2 = (4x)^2 +(3x)^2$
$100 = 16x^2 + 9x^2$
$100 = 25x^2$
$x^2 = 4$
$x = 2$ , $" so " 4x = 8$ , $3x = 6$
This means from $D$ we need to go up 8, over 6 from $D$ to find $C (8, 4)$
Note: The location of $C$ depends on $D$ , which can vary. This means we don't know the length of $DA$ or $AB$ .
Find the $m_(BC ) = (y_2 - y_1)/(x_2 - x_1) = (8 - 4)/(5-8) = 4/-3 = -4/3$ $" so " BC " is perpendicular to " AB$

So $ABCD = "rectangle"$
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