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Andrew [12]
3 years ago
7

For each pair of points below, think about the line that joins them. For which pairs is the line parallel to the y-axis? Circle

your answer(s). Then, give 2 other coordinate pairs that would also fall on this line.
a. (3.2, 8.5) and (3.22, 24)
b. (13 1/3, 4 2/3) and (13 1/3, 7)
c. (2.9, 5.4) and (7.2, 5.4)
Mathematics
1 answer:
gogolik [260]3 years ago
5 0

Answer:

Option B.

Step-by-step explanation:

If two lines are parallel then their slopes are always same.

Following this rule we can find the slope by the given pairs of coordinates of the options.

If the slope of the line is same as the slope of y axis then the line passing through these points will be parallel to the y axis.

Slope of y - axis = ∞

Option A). Slope = \frac{y-y'}{x-x'}

                            =  \frac{24-8.5}{3.22-3.2}

                            = \frac{15.5}{0.02}

                            = 775

Therefore, line passing through points (3.2, 8.5) and (3.22, 24) is not parallel to y axis.

Option B). Slope of the line passing through (\frac{40}{3},\frac{14}{3}) and (\frac{40}{3},7) will be

= \frac{\frac{14}{3}-7}{\frac{40}{3}-\frac{40}{3}}

= ∞

Therefore, line passing though these points is parallel to the y axis.

Option C). Slope of the line passing through (2.9,5.4) and (7.2, 5.4)

= \frac{5.4-5.4}{7.2-2.9}

= 0

Therefore, slope of this line is not equal to the slope of y axis.

Option B. is the answer.

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neonofarm [45]
Answer:

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see graph below

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First, we need to state the coordinates of the original image:

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\begin{gathered} Dilation\text{ rule:} \\ (x,\text{ y\rparen}\rightarrow(kx,\text{ ky\rparen} \\ where\text{ k = scale factor} \\  \\ scale\text{ factor = 1/2} \\ M^{\prime}\text{ = \lparen}\frac{1}{2}(3),\text{ }\frac{1}{2}(-2)) \\ M^{\prime}\text{ = \lparen}\frac{3}{2},\text{ -1\rparen} \\  \\ F\text{ = \lparen}\frac{1}{2}(4),\text{ }\frac{1}{2}(-2)) \\ F^{\prime}\text{ = \lparen2, -1\rparen} \end{gathered}\begin{gathered} L\text{ = \lparen}\frac{1}{2}(1),\text{ }\frac{1}{2}(-5)) \\ L^{\prime}\text{ = \lparen}\frac{1}{2},\text{ }\frac{-5}{2}) \\  \\ W\text{ = \lparen}\frac{1}{2}(5),\text{ }\frac{1}{2}(-5)) \\ W^{\prime}\text{ = \lparen}\frac{5}{2},\text{ }\frac{-5}{2}) \end{gathered}

The new coordinates:

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Plotting the coordinates:

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