<p>Hi!
Thanks to anyone who can help me out with this!</p>
<p>These are problems for the Jan SAT test.
I'm doing a review and I can't figure them out...</p>
<p>7) In the xy-plane, the points (1,-3), (2,2,) and (3,p) are collinear. What is the value of P?</p>
<p>(A) 1
(B) 3
(C) 4
(D) 7
(E) 8</p>
<p>I got the right answer (which is D, but I have no clue how I got it right... my guessing skills were top-notch I guess lolz.)
So can anyone tell me the process?</p>
<p>8) if y=x^2-1 and x is a real number, which of the following CANNOT be the value of y?</p>
<p>(A) -2
(B) -1
(C) 0
(D) 1
(E) 2</p>
<p>I got this wrong. Can anyone tell me the process? </p>
<p>thanks ^__^</p>
<p>7) If the points are collinear, it means that they all lie on the same line. Therefore, the slope of the line is constant, no matter which two points you choose. So using the first two points, we see that the slope of the line is (2-negative3)/(2-1) = 5. That means that (p-2)/(3-2) also must equal 5. Thus, p=7, which is choice D.</p>
<p>8) This problem is basically a reasoning test to see if you realize that x^2 always is going to be a positive value, assuming x is a real number. So first, isolate x^2 on one side of the equation. Thus: x^2 = y+1. Now, you can simply plug in the answers choices and see which value of y makes (x^2) not positive. Choices C, D, and E all make x^2 either 1, 2, or 3, respectively. That’s fine because x^2 is still positive. Now try choice B. If y = -1, then x^2 = -1+1, since x^2 = y+1. That’s also fine, because if x^2 = 0, then x must be 0, which is a real number. However, if you try choice A, you realize that x^2 = -2+1 = -1. x^2 can’t possibly be -1 if x is a real number. So the answer is -2 (Choice A). </p>
<p>Hope that helps!</p>
<p>8) x^2 >= 0
<=> x^2 - 1 >= -1 > -2
<=> y > -2
=> answer A</p>
<p>^^Thank you!
The way you explained it helped me to understand 8. :)</p>
<p>But, 7 still has me confused.
I understand the idea that collinear means that it must be in a straight line.
But using what you wrote
" (2-negative3)/(2-1) = -5."
I redid the problem and plugged in (2-negative3) / (2-1), which is the same thing you did but I got the answer positive 5 which makes the answer A, which I know is wrong. Do you know what I might be doing wrong?</p>
<p>Thank you for helping me out! ^_^</p>
<p>^^
That makes a lot of sense. I wouldn’t have thought of it like that at all.</p>
<p>Thank you. ^_^</p>
<p>7) i’ll most probably make a sketch (which’ll take no more than 20s), but if you want an arithmetic solution, here you go:
let the general equation of the line containing the 3 points be y = ax + b. plug in, we have a set of two equations:
a + b = -3
2a + b = 2
=> a = 5; b = -8
=> y = 5x - 8
plug in => p = 3*5 - 8 = 7
hence (D).</p>
<p>7) In the xy-plane, the points (1,-3), (2,2,) and (3,p) are collinear. What is the value of P?</p>
<p>(A) 1
(B) 3
(C) 4
(D) 7
(E) 8</p>
<p>It’s a linear line, so the standard formula is y = ax + b
Then a (the slope) = (2 - -3) / (2 - 1) = 5
We have y = 5x + b.
We don’t care about b, since we just work from point (2,2). 1 (x) forward means 5 (y) upward. 2 + 5 = 7 = Answer D.</p>
<p>7) In the xy-plane, the points (1,-3), (2,2,) and (3,p) are collinear. What is the value of P?</p>
<p>(A) 1
(B) 3
(C) 4
(D) 7
(E) 8</p>
<p>Here’s a very quick and easy way to do this problem:</p>
<p>Since the points lie on the same line, there is a constant slope. In English this means that equal “jumps” in x yield equal “jumps” in y. </p>
<p>The x-coordinates are 1, 2, 3. Thus, the y-coordinates are -3, 2, 7. So p=7, choice (D).</p>
<p>To clarify, the “jumps” in x are equal - they are each 1 unit apart. Therefore the “jumps” in y must be equal. Since you add 5 to get from -3 to 2, you must add 5 to get from 2 to p. So p must be 7.</p>
<p>Let me know if you need further clarification.</p>
<p>Thanks a lot!!!
This really helped me out!
^_^</p>