Geometry Problems of the Day (Geometry Regents, January 2013)

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 2013

Part II: Each correct answer will receive 2 credits. Partial credit is possible.


32. Write an equation of a circle whose center is (3,2) and whose diameter is 10

Answer:

Remember to flip the sign of the center because the formula has MINUS signs in it. Change the diameter into a RADIUS and then square that.

(x + 3)² + (y - 2)² = 25

There isn't any work to show, but feel free to write "10/2 = 5, 5² = 25" if you want to be sure. Knowing the formula and where to put all the numbers is what is required.

32. Using a compass and straightedge, construct a line perpendicular to line ℓ through point P.
[Leave all construction marks.

Answer:

From P make an arc that intersects line L twice. From each of those two points, make the same size arc on the other side of line L from where P is. Where those two arcs intersect, label it point Q. Use the straightedge and draw line PQ.

34. Write an equation of the line that is the perpendicular bisector of the line segment having endpoints (3,-1) and (3,5). [The use of the grid below is optional.]

Answer:

Find the slope of the given line segment. Then find the perpedicular slope. Next, find the midpoint of the given line segment. Finally, write the equation of the line with the perpendicular slope that goes through that point.

The slope of the line between (3, -1) and (3, 5) is undefined. It's vertical. This makes life a lot easier. The perpendicular line will be horizontal, of the form y = b.

The midpoint of (3, -1) and (3, 5) is (3, 2). The perpendicular line through (3, 2) is y = 2.

More to come. Comments and questions welcome.

More Regents problems.

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