Pearson Geometry Common Core, 2011
PG
Pearson Geometry Common Core, 2011 View details
2. Perpendicular and Angle Bisectors
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Exercise 39 Page 299

Start by drawing AB and plotting the given points on a coordinate grid.

C

Practice makes perfect

We are asked which of the given points lies on the perpendicular bisector of AB. Let's start with a diagram.

We can see above that neither (3,12), nor (1,5), nor (3,3) are on the perpendicular bisector of AB. Hence, the only point which might be on the perpendicular bisector of AB is (6,6). To determine so, let's recall the Converse of the Perpendicular Bisector Theorem.

Converse of the Perpendicular Bisector Theorem

If a point is equidistant from the endpoints of a segment, then it is on the perpendicular bisector of the segment.

Therefore, to determine if (6,6) is on the perpendicular bisector of AB, we will find the distance to the endpoints of AB. To do so, we will use the Distance Formula. d=sqrt((x_2-x_1)^2+(y_2-y_1)^2) Let's find the distance between A(1,3) and (6,6) by substituting (1,3) and (6,6) for (x_1,y_1) and (x_2,y_2) in the above formula.
d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)
d=sqrt(( 6- 1)^2+( 6- 3)^2)
â–Ľ
Simplify right-hand side
d=sqrt(5^2+3^2)
d=sqrt(25+9)
d=sqrt(34)
We found that the distance between A(1,3) and (6,6) is sqrt(34). Let's now find the distance between B(1,9) and (6,6) using substitution one more time.
d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)
d=sqrt(( 6- 1)^2+( 6- 9)^2)
â–Ľ
Simplify right-hand side
d=sqrt(5^2+(- 3)^2)
d=sqrt(25+9)
d=sqrt(34)
We found that the distance between B(1,9) and (6,6) is also sqrt(34). Therefore, (6,6) is equidistant from the endpoints of AB and thus it is on the perpendicular bisector of AB. Hence the correct choice is C.