assignment 16

Good work.

Let me know if anything is unclear, and include specifics about what you do and do not understand.

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00:32:03 **** query 2.1.28 (was 2.1.18). Dist (a, a) to (0, 0).

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RESPONSE --> we would use distance formula (1) d=sqrt[0-a]^2+(0-a)^2 d=sqrt a^2+a^2 d=sqrt a^4

a^2 + a^2 = 2 a^2, not a^4.

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00:34:09 ** 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 --> I understand now

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00:41:00 **** query 2.1.22 (was 2.1.12). Dist (2,-3) to (4,2).

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RESPONSE --> using the distance formula we get d=sqrt[4-2]^2+(2-(-3))^2 d=sqrt12+13 d=sqrt25 d=5

d=sqrt[4-2]^2+(2-(-3))^2 = sqrt( 2^2 + 5^2 ) = sqrt(4 + 25) = sqrt(29).

This would be the correct solution to the problem. Not sure where you got 12 and 13.

The given solution below is incorrect; some of the numbers got transposed there.

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00:46:21 ** 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 --> using the distance formula we get d=sqrt[4-2]^2+(2-(-3))^2 d=sqrt12+13 d=sqrt25 d=5 in the book points are p1 (2,-3) and p2 (4,2) so x2-x1 would be 4-2 not 2-4 like you have in your answer or I am completly confused one

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01:39:24 **** query 2.1.30 (was 2.1.20). (-2, 5), (12,3), (10, -11) A , B, C.

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RESPONSE --> find lenght of each side by using distance formula d(a,b)=sqrt[12-(-2)]^2+(3-5)^2=sqrt196+4=sqrt200 d(b,c)=sqrt[10-12]^2+(-11-3)^2=sqrt4+196=sqrt200 d(a,c)=sqrt[10-(-2)]^2+(-11-5)^2=sqrt144+256=sqrt400=20 now we prove that it is a right triangle (sqrt200)^2+(sqrt200)^2=20^2 200+200=400 400=400 it is a right triangle now we find the area using the area of a triangle formula A=1/2bh A=1/2sqrt200(sqrt200)=1/2*200=100 square units 100 square units is the answer

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01:41:45 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 --> I guess I thought we were supose to find the area as well.

You were, and you did that correctly. Great work.

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22:12:37 **** query 2.1.46 (was 2.1.36) midpt btwn (1.2, 2.3) and (-.3, 1.1)

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RESPONSE --> using the midpoint formula we get. x=1.2+(-3)/2=-.9 and y=2.3+1.1/2=1.7 so the midpoint is (.9,1.7)

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22:13:17 ** 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 --> O.k.

The given solution is for points (1.2, 2.3) and (-3, 1.1). Missed the decimal in front of the 3.

The correct solution would have x coordinate x = ( 1.2+(-3) )/2 = -.9 / 2 = -.45. You didn't divide by 2.

The correct solution would be (-.45, 1.7).

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22:20:43 **** 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 --> an equilateral triangle means all three side are equal. so the vertex would be (2,4) or (2,-4)

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22:21:23 ** 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 --> o.k.

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22:28:06 **** What are the coordinates of the third vertex and how did you find them?

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RESPONSE --> (2,4) or (2-4) found them by knowing that an equalaterial triangle has three equal sides and we know that one side is 4 long so the other two must be the same also the two points given are a straight line so the vertex needs to be in beween (0,0) and (0,4) so it would be 2 but it would depend wich way the triangle was pointing whether it was a positive 4 or negetive 4.

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