Pearson Geometry Common Core, 2011
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Pearson Geometry Common Core, 2011 View details
Chapter Review
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Exercise 45 Page 424

The given figure is a parallelogram, so both pairs of opposite sides are parallel.

(a-b,c)

Practice makes perfect

We want to determine the coordinates of P. We will do that by first determining the coordinates of O. Then we will use that for a parallelogram both pairs of opposite sides are parallel.

Coordinates of P

First, notice that O is the origin. Therefore, its coordinates are (0,0). As arbitrary placeholders, let's label the coordinates of P as (x,y). We can call the other two points R and Q.

We know that the given figure is a parallelogram, so both pairs of opposite sides are parallel. Let's find the slopes of the sides by substituting the coordinates of the points into the Slope Formula.
Side Slope Formula Simplify
QR c-0/- b-(- a) c/- b+a
RP y-c/x-(- b) y-c/x+b
PO 0-y/0-x y/x
OQ 0-0/- a-0 0

Now we can use logic to find the values of x and y in terms of the given variables. Since RP and OQ are parallel, their slopes must be equal. y-cx+b= 0 In order for this fraction to equal 0, the numerator must be 0. y-c=0 ⇔ y= c Now we know that the y -coordinate of P is c. Since QR and PO are parallel, their slopes are also equal. Additionally, we can replace the y with c in the slope of PO. c/- b+a= y/x ⇔ c/- b+a= c/x Because the numerators are now the same, these slopes will be equal only if the denominators are equal. - b+a=x ⇔ x= a-b This tells us the x-coordinate of P. Finally, we have that the coordinates of point P are ( a-b, c).