<p>Hey buddy I didn’t really understand every word that you said. You are wrong about second one. First I said B too but it’s A because it doesn’t want us to find center of gravity.
It wants us to find center. There is 2 ways to find it
Try choice A
Draw a triangle. Draw any spot in it.
If we think the spot is (x,y)
They give us these (0,7) , (-2,3) , (2,3)
We chose two of them
(-2,3) , (2,3) -->for x it will be 0
(-2,3) , (0,7) -->for y
√( (0-(-2))^2 +(y-3)^2 ) = √( (0-0)^2 + (y-7)^2
y^2 -6y +4+9 = y^2 -14y+49
8y = 36
y = 4.5</p>
<p>Hahaha BS. Those are not from old SATs; if they are, they are pre-1995 or appeared in the experimental section. Number 3 has to have a typo. Number one should be seen as a composition of functions and the answer is plus/minus 1. H(x) can never be -2. Positive or negative one, however, will yield 2 and K(2) is always 0. K(-2) is 0 as well, but as I said -2 can never be attained from H(x).</p>
<p>Number two was easy. The answer is (0, 4.5) so A.</p>
<p>Those questions are NOT written by ETS. I don’t know what it is worse … the fact that PR is still unable to write decent and relevant questions or that a TEACHER does not better than that. Fwiw, it is yet another clear signal to AVOID getting your help or advice for the SAT within the four corners of your HS, unless it comes from one of your peers who has made the effort to … practice with real questions. </p>
<p>Not that it is important, but the Q2 is from one of PR’s facebook pages, namely The Princeton Review Lucknow. This might explain why the questions are horrendous.</p>
<p>Guy, #2 is very simple with an algebraic equation.
You know that the point has to be on the y axis since Amy and Claire live equidistant to the y-axis.
So, you let the distance from (0,3) to the point you are looking for be x. (0,3) because its is on the horizontal line from Amy to Claire.
Note that x, distance from amy/claire to the y axis (which is 2), are two legs of a right triangle.</p>
<p>Set up equation:</p>
<p>4-x = sqrt(x^2+2^2)</p>
<p>since those are the two distances to need to be the same. Then solve.</p>
<p>ugurduzel, you are wrong about no, 2: Orthocenter is equidistant from vertices, not the centroid. Easy mistake to make, I had to look it up myself. To find the answer for no 2, all you need to do is set up the equation in the isosceles triangle of sides 5, 5, and 6 perp bisector of 4, sqrt(x^2 +3^2) = 4-x because sqrt(x^2 +3^2) is the length of the line from one vertices to a point on the perp bisector at a diagonal, depending on the x (and let x stands for the distance up from the base), which will be equal to the length of the entire perpendicular bisector of the base minus the length going up the bisector you had to go up (because the point x you reach must be equidistant from the base vertices of triangle to the point on the bisector, and the point on the bisector to the top vertices). solve for x and you get that 8x=7 => x= 7/8 so add 3 (the distance from the origin to the base of the triangle) to the x and you get 3.785, not in answer choices. Conclusion, don’t listen to your teacher! :)</p>
<p>@MITer94 Oops, my mistake. You’re right. But the idea remains true: the center of gravity isn’t necessarily the point equidistant from the vertices. I think I switch the two up by mistake because the word “orthocenter” rolls off the tongue much smother and sounds cooler than the word “circumcenter.” But that’s just opinion. :D</p>