Wednesday, July 03, 2024

June 2024 Geometry Regents Part IV

This exam was adminstered in June 2024 .

June 2024 Geometry, Part IV

Each correct answer is worth up to 6 credits. Partial credit can be given. Work must be shown or explained.

35. Triangle JOE has vertices whose coordinates are J(4,6), O(-2,4), and E(6,0).
Prove that △JOE is isosceles.
[The use of the set of axes on the next page is optional.]

Point Y(2,2) is on OE.
Prove that JY is the perpendicular bisector of OE.


As far as Part IV questions go, this one isn't terrible. In fact, it's pretty simple compared to some of the ones they ask involving parallelograms.

To show that JOE is an isosceles triangle, you have to find the lengths of all three sides, state which two of them are congruent. Then state JOE is an isosceles triangle because it has to congruent sides. (YES! You need to write this concluding statement!)

You can use the Distance Formula, or, if you graph JOE, you can just count boxes and use Pythagorean Theorem.

JO = √(62 + 22) = √(40)

OE = √(82 + 42) = √(80)

EJ = √(22 + 62) = √(40)

Since JO ≅ EJ, triangle JOE is an isosceles triangle.

To prove that JY is the perpendicular bisector of OE, you have to prove that JY perpendicular to OE and that JY is the bisector of OE. Yes, you need to do both things.

If the lines are perpendicular, then the slopes will be inverse reciprocals (that is, they have a product of -1). If JY is a bisector, then it goes through the midpoint of OE.

The midpoint of OE is ( (-2+6)/2, (4+0)/2 ) or (2,2), which is point Y. So Y is the midpoint of OE and JY is a bisector of OE.

The slope of OE is (0-4)/(6-(-2)) = -4/8 = -1/2. The slope of JY is (6-2)/(4-2) = 4/2 = 2/1. The slopes are inverse reciprocals, and (-1/2)(2/1) = -1, so OE is perpendicular to JY.

Therefore JY is the perpendicular bisector of OE.

End of Part Exam

How did you do?

Questions, comments and corrections welcome.

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