# A rectangle ABCD is plotted on a graph paper in the first quadrant having A(9,2) , B(15,b) , C(a,13) , D(5,5). Find the value of a-b.

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If ABCD is a rectangle in the first quadrant, then the slopes of AB and CD must equal, and the slopes of BC and AD must equal. Also, the slopes AB and BC are negative reciprocals so the lines are perpendicular.

`m_{AB}={b-2}/{15-9}={b-2}/6`

`m_{CD}={5-13}/{5-a}={-8}/{5-a}=8/{a-5}`

when they are made equal, this gives the equation:

`{b-2}/6=8/{a-5}`

`(a-5)(b-2)=48`

`ab-5b-2a+10=48`

`ab-5b-2a=38` call this equation (1)

Also, comparing the other slopes, we get:

`m_{BC}={13-b}/{a-15}`

`m_{AD}={5-2}/{5-9}=-3/4`

making these equal, we get:

`{13-b}/{a-15}=-3/4`

`4(13-b)=-3(a-15)`

`52-4b=-3a+45`

`a={4b-7}/3` call this equation (2)

Sub (2) into (1) to get:

`{4b-7}/3b-5b-2{4b-7}/3=38` multiply by 3 and simplify

`4b^2-7b-15b-8b+14=100`

`4b^2-30b-100=0` divide by 2

`2b^2-15b-50=0` use quadratic formula

`b={15+-sqrt{225+4(2)(50)}}/4`

`={15+-25}/4` but in first quadrant so only positive root

`=40/4`

`=10`

Now sub into (2) to get:

`a={4(10)-7}/3=33/3=11`

**This means that `a-b=1` .**