Trouble with hard QAS math question

Test Preparation SAT Preparation
1

<p>Tough question from an old March QAS, does anyone have any easy way to solve it? It took me wayyy too long, because did the following: found the chord between two of the points, found the line perpendicular, did that for a different set of two points, and found where the lines intersect. Any easier way? Thanks :) The question is below:</p>

<p>In the coordinate plane, the points F (-2, 1), G (1, 4), and H (4, 1) lie on a circle with center P. What are the coordinates of point P?</p>

<p>A. (0,0)
B. (1,1)
C. (1,2)
D. (1, -2)
E. (2.5, 2.5)</p>

2

<p>Suppose that the coordinates of the center P are (x,y). Then the distance from the center to each of the three points is the same.</p>

<p>F (-2, 1), G (1, 4), and H (4, 1)</p>

<p>(x+2)^2 + (y-1)^2 = (x-1)^2 + (y-4)^2 = (x-4)^2 + (y-1)^2</p>

<p>Look at the first and third expressions, and note that the (y-1)^2 cancel out. The only way (x+2)^2 = (x-4)^2 is when x=1 (3<em>3 = (-3)</em>(-3)). Substitute x=1 in:</p>

<p>(x+2)^2 + (y-1)^2 = (x-1)^2 + (y-4)^2 to get (y-4)^2 - (y-1)^2 = 9. Solve to get y = 1.</p>

3

<p>wow thats a unique way to look at it, thanks!</p>

4

<p>Another way to do it is to start with choice (C) and see if the distances from (1,2) to each of the 3 points is the same.</p>

<p>For example, for the first point d^2 = 3^2 + 1^2 = 10
For the second point d^2 = 0^2 + 2^2 = 4.</p>

<p>So we can eliminate choice (C).</p>

<p>Now choice (B):</p>

<p>First point: d^2 = 3^2 + 0^2 = 9
Second point: d^2 = 0^2 + 3^2 = 9
Third point: d^2 = 3^2 + 0^2 = 9</p>

<p>Therefore the answer is choice (B).</p>

5

<p>You just don’t need coordinate geometry on the SAT. In fact, I’ve never used the distance formula on an SAT problem ever. </p>

<p>These problems are always designed with little “coincidences” built in that make them easier than they look at first. To discover the coincidences, you just need to draw a really neat diagram and then count boxes. Try drawing a graph – make it really neat. And then you will see right away that the third given point lies vertically above the midpoint of the other two and that all of them are 3 boxes away from that midpoint.
So that’s it. </p>

<p>You may complain that my method won’t always work. And you’d be right if this were school. But it’s not…</p>