Assignment 16

course Mth 158

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16:40:56

**** 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)

d = sqrt (a^2 + a^2)

d = sqrt(2a^2)

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16:41:03

** 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 -->

ok

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16:42:16

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

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RESPONSE -->

d = sqrt((2.3 - -2)^2 + (2 - -3)^2)

d = sqrt(4 + 25)

d = sqrt(29)

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16:44:01

** 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 -->

Why do you use the x1 and y1 first in the equaitons. Isn't it suppose to be the x2 -x1 etc.

The error in the given solution is the 2 - 4 is not 4, but 2.

The choice of which point is 'point 1' and which is 'point 2' is arbitrary; however the order you used is more conventional and more in keeping with the way the problem is stated.

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16:48:57

**** 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) = 10*sqrt(2)

d(A,B) = sqrt((10 - -12)^2 + (-11 - 3)^2) = 10*sqrt(2)

d(A,B) = sqrt((10 - -2)^2 + (-11 - 5)^2) = sqrt(400) = 20

(10*sqrt(2))^2 + (10*sqrt(2))^2 = 20^2

400 = 400

Triangle is a right triangle.

A = 1/2* (10*sqrt(2)) * (10*sqrt(2)) = 100 sq. units

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16:49:05

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 -->

ok

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16:50:25

**** 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

Midpoint is (.45,1.7)

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16:50:51

** 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 -->

The book has -.3 instead of -3

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16:51:15

**** 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 -->

Not sure how to do this.

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16:51:24

** 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 -->

ok

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16:51:39

**** What are the coordinates of the third vertex and how did you find them?

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RESPONSE -->

?

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Good work. Let me know if you have any questions on that last problem.