course mth164 ??Z???????????assignment #010010. Dot product, vector algebra
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08:07:54 `q001. The dot product of two vectors is equal to the product of the magnitudes of the vectors, multiplied by the cosine of the angle between the vectors. What is the dot product of vector v having magnitude 10 and angle 30 degrees, with vector w having magnitude 8 and angle 90 degrees?
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RESPONSE --> The angle between the two lines is 90-30=60 degrees Using the formula the dot product would be 8x10xcos(60) = 40 confidence assessment: 3
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08:08:03 The magnitudes are 10 and 8, and the angle between the vectors is the change in angle from 30 degrees to 90 degrees, or 60 degrees. The dot product is therefore dot product = product of magnitudes * cos(angle) = 10 * 8 * cos(60 deg) = 40.
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RESPONSE --> ok self critique assessment:
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08:11:47 `q002. What are the x and y components of the vector v having magnitude 10 and angle 30 degrees, and vector w having magnitude 8 and angle 90 degrees? What do you get if you add the product of the two x components to the product of the two y components? How is this result related to the answer to the preceding exercise?
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RESPONSE --> For vector v: sin30=y/10 so y = 5 cos30=x/10 so x=5sqrt(3) For vector w: x is 0 and y is 8 Multiplying components gives: 5sqrt(3)(0) + 5(8) = 40 This is the same answer as the dot product of the previous problem confidence assessment: 3
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08:12:00 The vector v has x component vx = 10 cos(30 deg) = 8.7, approx., and vy = 10 sin(30 deg) = 5. The vector w has x component wx = 8 cos(90 deg) = 0 and y component wy = 8 sin(90 deg) = 8. The product of the two x components is vx * wx = 8.7 * 0 = 0, and the product of the y components is vy * wy = 5 * 8 = 40. The sum of these products is 0 + 40 = 40, which is identical to the result of the preceding exercise in which we found the dot product of v and w.
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RESPONSE --> ok self critique assessment:
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08:27:03 `q003. If vector v is represented by < v1, v2 > and vector w by < w1, w2 > then if the result of the preceding exercise is valid, how do we write in symbols the dot product of the two vectors? In symbols how do we write the magnitudes of the two vectors? How then do we write the statement that the dot product of the two vectors is equal to the product of the magnitudes of the vectors, multiplied by the cosine of the angle between the vectors?
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RESPONSE --> The dot product =v1 x w1 + v2 x w2 |v| = sqrt(v1^2 + v2^2) |w|=sqrt(w1^2 + w2^2) v1xw1 + v2xw2 = |v||w|xcos theta confidence assessment: 2
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08:27:51 If the preceding exercise generalizes then the dot product is the sum of the product of the x components and the product of the y components. In this case we would therefore say that the dot product is v1 * w1 + v2 * w2. The magnitudes of the two vectors are | v | = sqrt(v1^2 + v2^2) and | w | = sqrt(w1^2 + w2^2). The statement therefore says that v1 * w1 + v2 * w2 = sqrt(v1^2 + v2^2) * sqrt(w1^2 + w2^2) * cos(theta).
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RESPONSE --> ok i see where I went wrong on the final part of the question self critique assessment:
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08:31:19 `q004. Use the result of the preceding exercise to find the cosine of the angle between the vectors < 2, 3 > and < -7, 4 >. What therefore is the angle between these vectors?
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RESPONSE --> cos(theta)=(2x-7 + 3x4)/(sqrt(13)sqrt(65) cos(theta) = -2/sqrt(845) theta = 93.9 degrees confidence assessment: 3
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08:32:36 The cosine of the angle theta between the vectors is found using the fact that dot product is product of magnitudes multiplies by the cosine of the angle, expressed in detail as v1 * w1 + v2 * w2 = sqrt(v1^2 + v2^2) * sqrt(w1^2 + w2^2) * cos(theta). We easily rearrange the equation to get cos(theta) = [ v1 * w1 + v2 * w2 ] / [ sqrt(v1^2 + v2^2) * sqrt(w1^2 + w2^2)]. In this case we have cos(theta) = [ 2 * -7 + 3 * 4 ] / [ sqrt(2^3 + 3^2) * sqrt( (-7)^2 + 4^2) ] = -2 / ( sqrt(13) * sqrt(53) ] = -2 / 26.3 = .08 approx.. The angle is therefore the solution theta to the equation cos(theta) = .08, which gives us theta = 83 degrees, approx..
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RESPONSE --> I got sqrt(65) for the magnitude of the second vector instead of sqrt(53) self critique assessment:
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08:42:02 `q005. The vectors v = < 2, 4, 5 > and w = < -3, 7, 2 > exist in 3-dimensional space. What do you think are the magnitudes of these two vectors?
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RESPONSE --> |v| = sqrt(2^2 + 4^2 + 5^2) = sqrt(45) |w| = sqrt((-3)^2 + 7^2 + 2^2) = sqrt (62) confidence assessment: 3
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08:42:57 The magnitude of a vector is found by taking the square root of the sum of the squares of its components, just as for a vector in 2-dimensional space. In this case we get magnitudes sqrt(2^2 + 4^2 + 5^2) = sqrt(45) = 6.7, approx., and sqrt((-3)^2 + 7^2 + 2^2) = sqrt(76) = 8.7 approx..
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RESPONSE --> I got the answer to the last vector magnitude as sqrt(62) instead of sqrt(76) self critique assessment:
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08:46:18 `q006. What is the dot product of the vectors v = < 2, 4, 5 > and w = < -3, 7, 2 > ? What therefore is the angle between these vectors?
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RESPONSE --> dot product = 2x-3 + 4x7 + 5x2 = 32 cos(theta) = 32/(sqrt(45)sqrt(62)) cos(theta)=.6058 theta = 52.7 degrees confidence assessment: 3
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08:47:08 Since dot product = product of magnitudes * cos(theta) we have cos(theta) = dot product / product of magnitudes. In the preceding problem we found the magnitudes of the vectors to be sqrt(45) and sqrt(76). So if we can find the dot product we can find the cosine of the angle and therefore the angle. The dot product is again the sum of the products of the invidual components, in this case 2 * (-3) + 4 * 7 + 5 * 2 = 32. Thus we have cos(theta) = 32 / ( sqrt(45) * sqrt(76) ) and theta = arccos[ 32 / (sqrt(45) * sqrt(76) ) = 56.8 deg, approx.. Note that these vectors can actually be constructed in 3-dimensional space, and if the construction is accurate the angle will be as indicated.
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RESPONSE --> Ok, my angle is off a little because I have the magnitude of vector w to be sqrt(62) self critique assessment:
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08:49:34 `q007. Calculate the angle between the two vectors < 1, 7, -3, 4 > and < -3, -5, 2, 7 > .
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RESPONSE --> cos(theta) = (1x-3+7x-5+-3x2+4x7)/(sqrt(75)sqrt(87)) cos(theta)=-16/sqrt(6525) theta = 101 degrees confidence assessment: 3
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08:49:40 As before we find that theta = arccos ( dot product / product of magnitudes ). In this case theta = arccos [ (1 * -3 + 7 * -5 + -3 * 2 + 4 * 7) / ( sqrt(1^2 + 7^2 + (-3)^2 + 4^2 ) * sqrt( (-3)^2 + (-5)^2 + 2^2 + 7^2 ) ] = arccos[ -16 / 80.8 ] = 101 degrees, approx.. Note that though these vectors occur in 4 dimensions and hence cannot be constructed in 3-dimensional space the calculation of the angle between them, extending in the most obvious manner the procedures of the preceding problems, give us a result with just a little additional calculation.
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RESPONSE --> ok self critique assessment:
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????U??????? assignment #011 011. Conic sections Precalculus II 07-11-2007
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12:14:10 `q001. What is the distance from (3, 5) to (7, 2)?
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RESPONSE --> 5 confidence assessment: 3
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12:14:16 The distance is the hypotenuse of a right triangle with vertices at the two points, whose legs are parallel to the x and y axes and whose hypotenuse runs from one point to the other. The legs of the triangle are thus 7 - 3 = 4 and 2 - 5 = -3, and the hypotenuse is sqrt(4^2 + (-3)^2) = sqrt(25) = 5.
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RESPONSE --> ok self critique assessment:
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12:14:45 `q002. What is the distance from (7, 2) to (x, y)?
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RESPONSE --> sqrt((7-x)^2+(2-y)^2) confidence assessment: 3
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12:14:50 The distance is the hypotenuse of a right triangle with vertices at the two points, whose legs are parallel to the x and y axes and whose hypotenuse runs from one point to the other. The legs of the triangle are thus x - 7 and y - 2, and the hypotenuse is sqrt((x-7)^2 + (y-2)^2).
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RESPONSE --> ok self critique assessment:
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12:15:13 `q003. What is the distance from (x1, y1) to (x, y)?
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RESPONSE --> sqrt((x1-x)^2+(y1-y)^2) confidence assessment: 3
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12:15:18 The distance is the hypotenuse of a right triangle with vertices at the two points, whose legs are parallel to the x and y axes and whose hypotenuse runs from one point to the other. The legs of the triangle are thus x - x1 and y - y1, and the hypotenuse is sqrt((x-x1)^2 + (y-y1)^2).
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RESPONSE --> ok self critique assessment:
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12:15:37 `q004. What is the distance from (x1, y1) to (x2, y2)?
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RESPONSE --> sqrt((x1-x2)^2+(y1-y2)^2) confidence assessment: 3
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12:15:42 The distance is the hypotenuse of a right triangle with vertices at the two points, whose legs are parallel to the x and y axes and whose hypotenuse runs from one point to the other. The legs of the triangle are thus x2 - x1 and y2 - y1, and the hypotenuse is sqrt((x2-x1)^2 + (yy-21)^2).
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RESPONSE --> ok self critique assessment:
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12:16:12 `q005. Write as an equation: The distance from (7, 2) to (x, y) is 9.
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RESPONSE --> 9=sqrt((7-x)^2+(2-y)^2) confidence assessment: 3
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12:16:33 The distance from (7, 2) to (x, y) is sqrt((x-7)^2 + (y-2)^2), so the statement says that sqrt((x-7)^2 + (y-2)^2) = 9. Note that both sides of this equation could be squared to get (x-7)^2 + (y-2)^2 = 81.
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RESPONSE --> ok I didn't simplify my equation by squaring both sides. self critique assessment:
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12:17:24 `q006. Write as a system of two equations: The distance from (7, 2) to (x, y) is 9 and distance from (4, 1) to (x, y) is 10. Solve the system for x and y.
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RESPONSE --> 81=(7-x)^2+(2-y)^2 100=(4-x)^2+(1-y)^2 confidence assessment: 3
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12:17:48 The distance from (7, 2) to (x, y) is sqrt((x-7)^2 + (y-2)^2), so the first statement says that sqrt((x-7)^2 + (y-2)^2) = 9. The distance from (4, 1) to (x, y) is sqrt((x-4)^2 + (1-y)^2), so the second statement says that sqrt((x-4)^2 + (1-y)^2) = 10.{} If we square both equations we get (x-7)^2 + (y-2)^2 = 81 and (x-4)^2 + (1-y)^2 = 100. Expanding the squares in these equations we get x^2 - 14 x + 49 + y^2 - 4 y + 4 = 81 and x^2 - 8 x + 16 + y^2 - 2 y + 1 = 100. Collecting terms we have{} x^2 - 14 x + y^2 - 4 y = 28 and x^2 - 8 x + y^2 - 2 y = 83. Subtracting the second equation from the first we get -6x - 2 y = -55, which we solve for y to get y = -3x + 55/2. Substituting this expression into the first equation we get {}x^2 - 14 x + 49 + (-3x + 55/2)^2 - 4 ( -3x + 55/2 ) + 4 = 81, which we expand to get 10?^2 - 167? + 701.25 = 83. Solving for x (using the quadratic formula) we get two solutions, x = 11.16 and x = 5.54. Substituting these x values into the second equation we get y^2 - 2? + 35.2656 = 83, with solution y = 7.98 or y = -5.98; and y^2 - 2? - 34071/2500 = 83 with solutions y = 10.88 and y = -8.88. This gives us possible solutions (11.16, 7.98), (11.16,-5.98), (5.54, 10.88) and (5.54, -8.88). {}Checking out these solutions with the first equation we see that (11.16, -5.98) and (5.54, 10.88) are in fact solutions, while (11.16, 7.98) and (5.54, 10.88) are not.
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RESPONSE --> ok I just wrote the systems of equations self critique assessment:
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12:24:46 `q007. Write as an equation: The distance from (7, 2) to (x, y) is equal to the distance from (4, 1) to (x, y). Simplify this equation.
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RESPONSE --> (7-x)^2+(2-y)^2=(4-x)^2+(1-y)^2 3x+y-18=0 confidence assessment: 3
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12:25:12 The distance from (7, 2) to (x, y) is sqrt((x-7)^2 + (y-2)^2), and the distance from (4, 1) to (x, y) is sqrt((x-4)^2 + (1-y)^2). To say that these distances are equal is to say that sqrt((x-7)^2 + (y-2)^2)= sqrt((x-4)^2 + (1-y)^2) Squaring both sides we get (x-7)^2 + (y-2)^2 = (x-4)^2 + (1-y)^2. Expanding the squares we x^2 - 14 x + 49 + y^2 - 4 y + 4 = x^2 - 8 x + 16 + y^2 - 2 y + 1. Subtracting x^2 - 8 x + 16 + y^2 - 2 y + 1 from both sides we get -6x - 2y + 36 = 0. Solving for y we get y = -3x + 18. This is a linear equation, telling us that the set of points (x, y) which are equidistant from (7, 2) and (4, 1) lie along a straight line with slope -3 and y-intercept 18. Recall from basic geometry that the perpendicular bisector of a line segment through two points is the line which is equidistant from those points. Since the slope of the segment from (7,2) to (4,1) is 1/3, we expect that the perpendicular bisector will have slope - 1 / (1/3) = -3, as is the case for the line we have obtained. It is also easy to verify that the line y = -3x + 18 contains the midpoint between (7, 2) and (4, 1): The midpoint is ( (7+4)/2, (2+1)/2 ) = (11/2, 3/2). Substitution will show that this point lies on the line y = -3x + 18.
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RESPONSE --> OK I didn't write my answer in slope/intercept form but it is the same. self critique assessment:
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12:27:32 `q008. Write as an equation: The distance from ((7, 4) to (x, y) is equal to the distance from the line y = 2 to (x, y). Simplify this equation.
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RESPONSE --> (7-x)^2+(4-y)^2=(0-x)^2+(2-y)^2 4y=-14x+61 or y=-7/2x+61/4 confidence assessment: 3
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12:28:07 The distance from (7, 4) to (x, y) is sqrt( (x-7)^2 + (y-4)^2 ). The distance from (x, y) to the line y = 2 lies along the vertical line from (x, y) to y = 2; it is clear from Figure 49 that this distance is | y - 2 |. Thus we have sqrt( (x-7)^2 + (y-4)^2 ) = y - 2. Squaring both sides we get (x-7)^2 + (y-4)^2 = (y - 2 ) ^ 2. Expanding the squares we have x^2 - 14 x + 49 + y^2 - 8 y + 16 = y^2 - 4 y + 4. Subtracting y&2 - 4 y + 4 from both sides we have x^2 - 14 x + 49 - 4 y + 12 = 0. You aren't expected to have known the rest of this solution before, but you need to note the following: Note that this equation is quadratic in x and linear in y. An equation of this form is generally rearranged into the form (y - k) = A * (x - h) ^ 2. In this case we can add 4y - 12 to both sides to get x^2 - 14 x + 49 = 4 ( y - 3). This simplifies to (y - 3) = 1/4 ( x - 7)^2.
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RESPONSE --> I missed the part about x^2 self critique assessment:
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12:36:16 `q009. Write as an equation: The distance from (7, 6) to (x, y) plus the distance from (7, 2) to (x, y) is 6. Simplify this equation.
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RESPONSE --> sqrt((7-x)^2+(6-y)^2)+sqrt((7-x)^2+(2-y)^2)=6 y^2-3x^2+16y+42x-158=0 confidence assessment: 2
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12:36:47 A sketch of the points (7, 2) and (7, 6), with a possible point (x,y) satisfying the description is shown in Figure 5. The distance from (x, y) to (7, 6) is (sqrt((x-7)^2 + (y-6)^2) and the distance from (x, y) to (7, 2) is sqrt((x-7)^2 + (y-2)^2). To say that the sum of these distances is 6 is to say that sqrt((x-7)^2 + (y-6)^2) + sqrt((x-7)^2 +(y-2)^2) = 6. This equation is a bit messy to simplify. If we first square both sides we get ((x-7)^2 + (y-6)^2) + 2 sqrt [((x-7)^2 + (y-6)^2) * ((x-7)^2 +(y-2)^2)] + ((x-7)^2 +(y-2)^2) = 36. We still have a square root in our equation so we rearrange the equation so that the square root is isolated on one side: 2 sqrt [((x-7)^2 + (y-6)^2) * ((x-7)^2 +(y-2)^2)] = 36 - ((x-7)^2 + (y-6)^2) - ((x-7)^2 +(y-2)^2) If we square both sides and are fortunate enough to keep all the variables straight we get 4?^4 - 112?^3 + 8?^2?^2 - 64?^2? + 1336?^2 - 112??^2 + 896?? - 7728? + 4?^4 - 64?^3 + 744?^2 - 3904? + 18020 = 4?^4 - 112?^3 + 8?^2?^2 - 64?^2? + 1192?^2 - 112??^2 + 896?? - 5712? + 4?^4 - 64?^3 + 664?^2 - 3264? + 10404, which looks pretty bad until we notice that the x^4, y^4, x^3, y^2, x^2 y^2, x^2 y, x y^2 and xy terms are the same on both sides so that if we simply subtract the entire right-hand side from both sides and the divide both sides by 16 we get 9?^2 - 126? + 5?^2 - 40? + 476 = 0. Note that this equation is quadratic in both x and y. An equation of this form does not always have real solutions, but when it does the graphed solution points will lie on either an ellipse or a hyperbola.
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RESPONSE --> I miss calculated somwhere through the expansions. self critique assessment:
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12:37:47 `q010. Write as an equation: The distance from (x, y) to (7, 6) is 6 units greater than the distance from (x, y) to (7, 2). Do not simplify this equation.
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RESPONSE --> sqrt((x-7)^2+(y-6)^2)=sqrt((x-7)^2+(y-2)^2)+6 confidence assessment: 3
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12:37:55 The distance from (x, y) to (7, 6) is sqrt((x-7)^2 + (y-6)^2) and the distance from (x, y) to (7, 2) is sqrt((x-7)^2 + (y-2)^2). To say that the first is 6 units greater than the second is to say that sqrt((x-7)^2 + (y-6)^2) - sqrt((x-7)^2 +(y-2)^2) = 6. This equation will end up being quadratic, as was the equation in the preceding exercise.
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RESPONSE --> ok self critique assessment:
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12:42:54 `q011. Sketch the points (7, 4) and (3, 2) on a set of coordinate axes. Locate the point halfway between these two points. Then locate three more points which lie at equal distances from the two points, but not on the line joining the two points.
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RESPONSE --> on the line (5,3) Not on the line: (3,2), (1,1) confidence assessment: 3
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12:43:02 This situation is depicted in Figure 22. The halfway point is found by simply averaging the x coordinates and averaging the coordinates: we get midway point: ( (7 + 3) / 2, (4 + 2) / 2) = (5, 3). Three other points are depicted, each showing the two equal segments (one from each of the original points to the equidistan point) which show that the points are equidistant. It is worth noting that these equidistant points all lie on the perpedicular bisector of the segment joining the two original points. You should be aware of how this is proved in terms of similar triangles which turn out to be right triangles.
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RESPONSE --> ok self critique assessment:
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12:49:29 `q012. Sketch the point (7, 4) on a set of coordinate axes, and also sketch the line y = 2 on the same set of axes. Sketch the point which lies halfway between the point and the line. Then locate a point which lies directly to the right of (7,4) which is at the same distance from (7,4) as from the line y = 2. Approximately locate also the points which are located as follows: 3 units above the line and also 3 units away from (7, 4) 4 units above the line and also 4 units away from (7, 4) 5 units above the line and also 5 units away from (7, 4).
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RESPONSE --> (9.8,5) would be 3 units above the line and 3 units from (7,4) (3.5,6) would be 4 units above the line and 4 units from (7,4) (3,16) would be 5 units above the line and 5 units from (7,4) confidence assessment: 2
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12:49:47 Figure 95 depicts the horizontal line y = 2 and the point (7, 4). The point halfway between (7, 4) and the line y = 2 is indicated; note that this point lies at (7, 3). Moving directly to the right of (7, 4) we encounter the indicated point directly to its right and equidistant from (7, 4) and the line y = 2. Note that the horizontal and vertical line segments from this point indicate these equal distances. We also see the points which lie 3 units from both (7,4) and the line y = 2, and 5 units from line and point; note for each of these points that the line segment to the point (7,4) is the same length as the segment directed to the y = 2 line.
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RESPONSE --> ok self critique assessment:
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12:50:45 `q013. Sketch the set of points (x, y) you think would satisfy the condition that the distance from (x, y) to (7, 4) is equal to the distance from the line y = 2.
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RESPONSE --> ok confidence assessment:
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12:51:09 In figure 15 we see the curve that passes through the indicated points, and which satisfies the given condition. Every point on this curve is equidistant from (7, 4) and from the line y = 2.
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RESPONSE --> ok self critique assessment:
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12:53:02 `q014. Sketch the points (7, 6) and (7, 2) on a set of coordinate axes. How far is the point (7, 7) from each of these points? What is the sum of these distances? What other point gives the same sum?
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RESPONSE --> From (7,6) it would be 1 unit in the vertical From (7,2) it would be 5 units in the vertical The sum is 6 Distance from (7,7) to (7,4) added distance from (7,7) to (7,10) confidence assessment: 3
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12:53:07 (7,7) lies directly above (7,6), at a distance of 1 unit. (7,7) lies exactly 5 units above (7,2). So the sum of the two distances is 1 + 5 = 6 units. The point (7, 1), lying 1 unit directly below (7,2), also has the characteristic that the sum of the two distances is 6.
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RESPONSE --> ok self critique assessment:
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12:57:26 `q015. We saw in the preceding exercise that (7, 7) lies at distance 1 from (7, 6) and at distance 5 from (7, 2), so that the sum of these distances is 6. What is the sum of the distances from (8, 4) to (7, 6) and to (7, 2)? What is the sum of the distances from (9, 4) to these points? What is the sum of the distances from (10, 4) to these points? Estimate the x coordinate that makes the sum of the distances from (x, 4) to (7, 6) and to (7, 2) equal to 6. If you can find the exact x coordinate required.
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RESPONSE --> From (8,4) the sum is 4.47 From (9,4) the sum is 5.67 From (10,4) the sum is 7.21 The x-coordinate could be 4.76 or 9.23 confidence assessment: 2
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12:57:34 In Figure 64 we see the indicated points. The figure shows that the points (8,4), (8,5) and (8,6) lie at increasing distances from (7,6) and from (7,2). The distance from (7,6) to (8,4) is sqrt( (8-7)^2 + (4-6)^2) = sqrt(1+4) = sqrt(5); the distance from (7,2) to (8,4) is also sqrt(5). So the sum of the two distances is 2 sqrt(5) = 4.46 approx., which is less than 6. The distance from (7,6) to (9,4) is sqrt( (9-7)^2 + (4-6)^2) = sqrt(4 + 4) = sqrt(8); the distance from (7,2) to (8,4) is also sqrt(8). So the sum of the two distances is 2 sqrt(8) = 5.56 approx., which is still less than 6. From the pattern of the preceding two points it should be clear that the sum of distances for the point (10,4) will exceed 6. To verify this we calculate the distance from (7,6) to (10,4) and see that the distance is sqrt(13), as is the distance from (7,2) to (10,4). Thus the sum of distances is 2 sqrt(13) = 7.2, approx., which does indeed exceed 6.
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RESPONSE --> ok self critique assessment:
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12:57:45 `q016. Sketch the set of all points (x, y) you think would satisfy the condition that the sum of the distances from (x, y) to (7, 6) and to (7, 2) is 6.
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RESPONSE --> ok confidence assessment:
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12:57:48 One such point will lie between (9,4) and (10,4); from the pattern of results in the preceding problem we expect that this point will lie much closer to (9,4) than to (10,4). Figure 29 shows the point C, as described. Points A and B also have the property that the sum of the distances to (7,6) and (7,2) is equal to 6. Figure 87 shows all these points with the curve every point of which whose property is the same as that of C, A and B. That is, for every point of this curve the sum of the distances from that point to (9,4) and to (10,4) is 6.
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RESPONSE --> ok self critique assessment:
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