-gre.ge.11 x,y of fourth corner of diagonal

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SUMMARY

The fourth vertex of the rectangle in the (x,y) coordinate plane is definitively (3,-7). This conclusion is drawn from the calculated deltas of the width and height, which are 3 and 2 respectively. By applying the coordinates of the existing vertices and confirming the relationships between the lines, it is established that the lines are not perpendicular, confirming the diagonal nature of the rectangle. Thus, the fourth corner is accurately determined without the need for complex equations.

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karush
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In the standard (x,y) coordinate plane below, 3 of the vertices of a rectangle are shown. Which of the following is the 4th vertex of the rectangle?

boyce_20201004144618.png

sorry about the huge image but couldn't find where to scale it down
a. (3,-7)
b. (4,-8)
c. (5,-1)
d. (8,-3)
e. (9,-3)

ok I don't think we need a bunch of equations to do this
$\delta$ x of the with is 3
$\delta$ y of the width is 2

so the fourth corner is
(6-3,-5-2)=(3,-7)
 
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I get that from (2, 1) to (6, -5) is a "delta" of (6- 2, -5- 1)= (4, -6). So the line parallel to that through (-1, -1) goes through (-1+ 4, -1- 6)= (3, -7) also.
 
Just to cover all bases I'd first show that the lines (-1, -1) to (6, -5) and (2, 1) to (-1, -1) are not perpendicular so the line from (-1, -1) to (6, -5) must be a diagonal.

-Dan
 
good point
by observation it sure looks like a rectangle
 

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