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course Mth 158

xvܹ|ysx{assignment #016

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v慬nؤ˺Zۋ

College Algebra

07-09-2006

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22:36:48

**** query 2.1.28 (was 2.1.18). Dist (a, a) to (0, 0).

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RESPONSE -->

Dist (a, a) to (0, 0)

use the distance formula...

distance = sqrt ( (x2-x1)^2+(y2 - y1)^2 )

substitute for the x's and y's..

sqrt((a-0)^2+(a-0)^2)

sqrt(a^2+a^2)

sqrt(2 a^2)

sqrt(2) * sqrt(x^2) = sqrt(2) * a.

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22:37:07

** 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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22:40:01

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

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RESPONSE -->

Dist (2,-3) to (4,2).

use the distance formula...distance = sqrt( (x2 - x1)^2 + (y2 - y1)^2 )

substitute in for the x's and y's using the initial problem..

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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22:40:08

** 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 -->

ok

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22:51:49

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

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RESPONSE -->

A= (-2, 5), B=(12,3) C=(10, -11)

Find the length of each side of the triangle by using the distance formula..

d(A,C)= sqrt((-2-10)^2 + (5+11)^2)

sqrt 144+256

sqrt 400

20

d(A,B)= sqrt((-2-12)^2+(5-3)^2)

sqrt(196+4)sqrt(200)

10 sqrt of 2

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

sqrt(4+196)

sqrt 200

10 sqrt of 2

This is a right trianlgle..

Pythagorean Theorem is..a^2 +b^2=C^2

(10sqrt2)^2+(10sqrt2)^2=(20)

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

100 * 2 + 100 * 2 = 400

200 + 200= 400.......or 400=400

so the theorem checks.

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22:52:32

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 -->

alright

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22:56:24

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

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RESPONSE -->

midpt btwn (1.2, 2.3) and (-.3, 1.1)

midpoint formula..( (x1 + x2) / 2, (y1 + y2) / 2)

substitute in the points into the formula..

((1.2-3)/2) ((2.3+1.1)/2)

add or subtract inside the paretheses first..

(-1.8 / 2) ( 3.4 / 2)

divide by 2 and get solution...

(-0.9, 1.7)

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22:56:39

** 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 -->

ok

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22:59:09

**** 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 midpoint of the coordinates (0,4) and (0,0) is (0,2).

This checks in that ( (x1 + x2) / 2, (y1 + y2) / 2)

( (0+0) / 2, (4+0) /2

(0,2)

The triangle has a length of 4, which must be met on all sides..

therefore, the third vertex lies to the left or to the right of the midpoint (0,2).. the y is 2, while the x is unknown.

take the distance from (0,0) to (x,2).

d= sqrt((x-0)^2 + (2+0)^2)

sqrt (x^2+2^2)

sqrt(x^2 + 2^2) = 4

square both sides to cancel out the sqrt..

x^2 + 2^2= 16

subtract 4 from both sides..

x^2 = 16 - 4

x^2= 12

x = +-sqrt(12)

+ or -sqrt 4 * sqrt 3 = + or - 2 * sqrt3

(2, -2 sqrt(3)) or (2, 2 sqrt(3)).. are the coordiantes for the third vertex.

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23:06:50

** 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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22:08:46

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

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RESPONSE -->

I believe I just did that in the last problem?

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Your work on this assignment is good. Let me know if you have questions.