Monday, August 23, 2021

Geometry Problems of the Day (Geometry Regents, June 2014)



Now that I'm caught up with the current New York State Regents exams, I'm revisiting some older ones.

More Regents problems.

Geometry Regents, June 2014

Part IV: A correct answer will receive 6 credits. Partial credit is possible.


38.The vertices of quadrilateral JKLM have coordinates J(-3,1), K(1,–5), L(7,-2), and M(3,4).

Prove that JKLM is a parallelogram.

Prove that JKLM is not a rhombus.

[The use of the set of axes below is optional.]


Answer:


You have a few options here. The graph can make it easier for you, but it isn't necessary to use.

To prove that JKLM is a parallelogram, you can show that the slopes of the opposite sides are equal, which makes them parallelogram. Two pairs of parallel sides makes a quadrilateral a parallelogram.

Slope of JK: (-5 - 1) / (1 - -3) = -6/4 = -3/2

Slope of KL: (-2 - -5) / (7 - 1) = 3/6 = 1/2

Slope of LM: (4 - -2) / (3 - 7) = 6/-4 = -3/2

Slope of MJ: (1 - 4) / (-3 - 3) = -3/-6 = 1/2

Opposite sides are parallel (same slope), therefore JKLM is a paralellogram. (You must state this! Don't just give a bunch of slopes and let the person scoring the question to mark you incomplete.)

You can also show that a quadrilateral is a parallelogram if the diagonals bisect each other. This is even to show. Find the midpoint of each diagonal. If they are the same point, then the diagonals bisect each other.

Midpoint of JL ( (-3 + 7)/2 , (1 + -2) / 2 ) = (2, -1/2)

Midpoint of KM ( (1 + 3)/2 , (-5 + 4) / 2 ) = (2, -1/2)

Since the midpoint of diagonals JL and KM is the same point (2, -1/2), the diagonals bisect each other. Therefore, JKLM is a parallelogram.

Finally, you could also find the length of the opposite sides. If they are equal, it is a parallelogram. That would require using the Distance Formula four times. The previous two methods are quicker and easier.




To show that JKLM is NOT a rhombus, you need to find a property that makes a parallelogram a rhombus and show that it isn't true. There are TWO that are easy to show. First, you could find the lengths of two consecutive sides. If they are not the same length, then it is not a rhombus. OR you could find the slopes of the diagonals because the diagonals of a rhombus are perpendicular to each other.

Method 1: (Note that you found the distances when did the slope. You can write it out again.)

Length of JK: SQRT(62 + 42) = SQRT(36 + 16) = SQRT(52)

Length of KL: SQRT(32 + 62) = SQRT(9 + 36) = SQRT(45)

Since JK is not congruent to KL, JKLM is not a rhombus.

Method 2:

Slope of JL: (-2 - 1) / (7 - -3) = -3/10

Slope of KM: (4 - -5) / (3 - 1) = 9/2

Multiply (-3/10) * (9/2) = -27/20. The product of the slopes is not -1, so the diagonals are not perpendicular. Therefore, JKLM is not a rhombus.

If you want to use the provided graph, you still need to prove the statements above. Just plotting the points and drawing the figure will not get you any credit. However, it might make it easier for you to find the slopes of the lines because you can count boxes. This also helps for showing Method 2, above.




End of Exam.

More to come. Comments and questions welcome.

More Regents problems.

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