Assignment 16

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13:00:17 **** query 2.1.28 (was 2.1.18). Dist (a, a) to (0, 0).

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RESPONSE --> d = sqrt( (0-a)^2 + (0-a)^2 ) = sqrt( -a^2 + -a^2) = sqrt(a + a)

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13:01:01 ** Using the distance formula we get distance = sqrt( (x2 - x1)^2 + (y2 - y1)^2 ) = sqrt((a-0)^2+(a-0)^2) = sqrt(a^2+a^2) = sqrt(2 a^2) = sqrt(2) * sqrt(x^2) = sqrt(2) * a. COMMON ERROR: sqrt(a^2 + a^2) = a + a = 2 a INSTRUCTOR'S CORRECTION: sqrt( x^2 + y^2 ) is not the same thing as x + y. For example sqrt(3^2 + 4^2) = sqrt(9 + 16) = sqrt(25) = 5 but 3 + 4 = 7. So you can't say that sqrt(a^2 + a^2) = a + a. **

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RESPONSE --> a^2 + a^2 cannot equal a + a

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13:03:37 **** query 2.1.22 (was 2.1.12). Dist (2,-3) to (4,2).

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RESPONSE --> d = sqrt( (4-2)^2 + (2 - (-3))^2 ) = sqrt( 2^2 + 5^2 ) = sqrt( 4+25) = sqrt(29) = 5.385

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13:04:35 ** using the distance formula we get distance = sqrt( (x2 - x1)^2 + (y2 - y1)^2 ) = sqrt((2-4)^2+(-3-2)^2) = sqrt((-4)^2+(-6)^2) = sqrt(16+36) = sqrt(52) = sqrt(4) * sqrt(13) = 2 sqrt(13) **

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RESPONSE --> I think either I wrote the problem down wrong when I worked it out or something. But what I did I think I got right.

There's a typo in the given solution; your solution is correct.

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13:15:41 **** query 2.1.30 (was 2.1.20). (-2, 5), (12,3), (10, -11) A , B, C.

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RESPONSE --> d(A,B) = sqrt( (12-(-2))^2 + (3-5)^2 ) = sqrt ( 14^2 + (-2)^2 ) = sqrt (196 + 4) = sqrt (200) = 14.14 d(B,C) = sqrt( (10-12)^2 + (-11+3)^2 ) = sqrt ((-2)^2 + (-8)^2 ) = sqrt ( 4 + 64) = sqrt (68) = 8.246 d(A,C) = sqrt( (10-(-2)^2 + (-11-5)^2 ) = sqrt ( 2^2 + (-16)^2 ) = sqrt (144+256) = sqrt (400) = 20

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13:19:59 STUDENT SOLUTION: The triangle is a right triangle if the Pythagorean Theorem holds. d(A,B)= sqrt((-2-12)^2+(5-3)^2) sqrt(196+4)sqrt(200) 10 sqrt2 d(B,C)= sqrt((12-10)^2+(3+11)^2) sqrt(4+196) sqrt200 10 sqrt2 d(A,C)= sqrt((-2-10)^2 + (5+11)^2) sqrt(144+256) sqrt(400) 20 The legs of the triangle are therefore both 10 sqrt(2) while the hypotenuse is 20. The Pythagorean Theorem therefore says that (10sqrt2)^2+(10sqrt2)^2=(20)^2 which simplifies to 10^2 (sqrt(2))^2 + 10^2 (sqrt(2))^2 = 20^2 or 100 * 2 + 100 * 2 = 400 or 200+200=400 and finally 400=400. This verifies the Pythagorean Theorem and we conclude that the triangle is a right triangle. **

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RESPONSE --> It looks like when you plug them in you are doing it backwards than I am. In the book, on page 160 it shows that you do this p = (x1, y1) and p = (x2, y2) thus giving: d = sqrt ( (x2 - x1)^2 + (y2 - y1)^2 ) you are doing (x1 -x2)^2 + (y1 - y2)^2 I am just doing what the book instructed. Sorry for the confusion.

Since the quantities are squared it makes no difference, but the order you used is more conventional and you should continue to use that order.

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13:27:15 **** query 2.1.46 (was 2.1.36) midpt btwn (1.2, 2.3) and (-.3, 1.1)

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RESPONSE --> x = ( 1.2 + (-.3) / 2) = .9 / 2 = .45 y = (2.3 + 1.1 / 2) = 3.4 / 2 = 1.7

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13:29:53 ** The midpoint is ( (x1 + x2) / 2, (y1 + y2) / 2) = ((1.2-3)/2) , ((2.3+1.1)/2) = (-1.8 / 2 , 3.4 / 2) = (-0.9, 1.7) **

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RESPONSE --> I got this one correct except for the x according to your notes. It looks like though that when you subtracted you did 1.2 + (-3) instead of 1.2 + ( -0.3). If you used the -0.3 it equals .9 if you use -3 it gives you -1.8. I think you might have wrote the problem down wrong. Sorry.

again you were right; the book changed the problem in the new edition and my solution doesn't match the stated problem

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13:29:53 ** The midpoint is ( (x1 + x2) / 2, (y1 + y2) / 2) = ((1.2-3)/2) , ((2.3+1.1)/2) = (-1.8 / 2 , 3.4 / 2) = (-0.9, 1.7) **

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RESPONSE -->

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13:40:17 **** query 2.1.50 (was 2.1.40). Third vertex of equil triangle with vertices (0, 0) and (0, 4).

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RESPONSE --> The third has to be (4,0) to make it equal but I am not sure how to work it out or even if my assumption is correct.

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13:41:29 ** The point (0, 2) is the midpoint of the base of the triangle, which runs from (0,0) to (0, 4). This base has length 4, so since the triangle is equilateral all sides must have length 4. The third vertex lies to the right or left of (0, 2) at a point (x, 2) whose distance from (0,0) and also from (0, 4) is 4. The distance from (0, 0) to (x, 2) is sqrt(x^2 + 2^2) so we have sqrt(x^2 + 2^2) = 4. Squaring both sides we have x^2 + 2^2 = 16 so that x^2 = 16 - 4 = 12 and x = +-sqrt(12) = +-sqrt(4) * sqrt(3) = +-2 * sqrt(3). The third vertex can therefore lie either at (2, 2 sqrt(3)) or at (2, -2 sqrt(3)). **

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RESPONSE --> First take midpoint then work out. I see how to work it out now.

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13:42:09 **** What are the coordinates of the third vertex and how did you find them?

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RESPONSE --> I didn't find them, but after reviewing notes I know how to now.

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Good work. Let me know if you have questions.