# A line passes through #(4 ,3 )# and #(1 ,4 )#. A second line passes through #(7 ,8 )#. What is one other point that the second line may pass through if it is parallel to the first line?

By the given condition of the problem the 2nd line is parallel to the !st one . So their slope will be equal.

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If the second line is parallel to the first line, it means that the slopes of the two lines are equal. The slope of the first line can be calculated using the formula:

[m = \frac{{y_2 - y_1}}{{x_2 - x_1}}]

where ((x_1, y_1)) and ((x_2, y_2)) are two points on the line.

For the first line: [(4, 3)] and [(1, 4)]

[m = \frac{{4 - 3}}{{1 - 4}} = \frac{{1}}{{-3}} = -\frac{{1}}{{3}}]

Since the second line is parallel to the first, it must also have a slope of (-\frac{{1}}{{3}}). Using the point-slope form of a line, we can find another point on the second line. Let's use the point ((7, 8)) which is given.

The point-slope form of a line is:

[y - y_1 = m(x - x_1)]

Substitute (x_1 = 7), (y_1 = 8), and (m = -\frac{{1}}{{3}}):

[y - 8 = -\frac{{1}}{{3}}(x - 7)]

[y - 8 = -\frac{{1}}{{3}}x + \frac{{7}}{{3}}]

[y = -\frac{{1}}{{3}}x + \frac{{7}}{{3}} + 8]

[y = -\frac{{1}}{{3}}x + \frac{{7}}{{3}} + \frac{{24}}{{3}}]

[y = -\frac{{1}}{{3}}x + \frac{{31}}{{3}}]

So, one other point that the second line may pass through if it is parallel to the first line is ((10, 21)).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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