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The quadrilateral ABCD, where A is (4,5) and C is (3,-2), is a square. Find the...

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nmeisu | Student, Grade 11 | Honors

Posted February 10, 2013 at 1:43 AM via web

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The quadrilateral ABCD, where A is (4,5) and C is (3,-2), is a square. Find the coordinates of B and D and hence or otherwise the area of the square.

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embizze | High School Teacher | (Level 1) Educator Emeritus

Posted February 10, 2013 at 2:47 AM (Answer #1)

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Given a square ABCD with the coordinates of A(4,5) and the coordinates of C(3,-2) find the area of the square:

The length of the diagonal of a square is `sqrt(2)` times the length of a side.

Using the distance formula we find `AC=sqrt((4-3)^2+(5-(-2))^2)`

so `AC=sqrt(50)=5sqrt(2)` .

The length of a side of the square is 5 so the area of the square is 25 square units.

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If you really want to know the coordinates of B and D:

`bar(BD)` is the perpendicular bisector of `bar(AC)` .

(a) Since it bisects `bar(AC)` we find the midpoint to be `((4+3)/2,(5-2)/2)=(3.5,3.5)`

(b) The slope of `bar(AC)` is `m=(5-(-2))/(4-3)=7` so the slope of the perpendicular bisector is `m=-1/7`

(c) The equation of the perpendicular bisector is `y-3.5=-1/7(x-3.5)` or `y=-1/7x+4`

(d) B and D will be the intersections of the line `bar(BD)` with a circle centered at (4,5) (or (3,-2) if you prefer) with a radius of 5.

The line: `y=-1/7x+4` so `-7y=x-28==>x=-7y+28`

The circle: `(x-4)^2+(y-5)^2=25` substituting for x we get:

`(-7y+24)^2+(y-5)^2=25`

`49y^2-336y+576+y^2-10y+25=25`

`50y^2-346y+576=0`

So `y=(173+-sqrt(1129))/50`

Substituting for y to get x the approximate coordinates for B are `(8.484,2.788)` and for D `(-0.924,4.132)` if teh vertices are read clockwise.

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