# A line passes through #(8 ,2 )# and #(6 ,7 )#. A second line passes through #(3 ,8 )#. What is one other point that the second line may pass through if it is parallel to the first line?

Infinite points given by

graph{((x-8)^2+(y-2)^2-0.02)((x-6)^2+(y-7)^2-0.02)((x-3)^2+(y-8)^2-0.02)(5x+2y-31)(5x+2y-44)=0 [-6.38, 13.62, 0, 10]}

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To find a point that the second line may pass through if it is parallel to the first line, we need to use the concept that parallel lines have the same slope.

The slope of the first line passing through (8, 2) and (6, 7) is calculated as:

( m_1 = \frac{y_2 - y_1}{x_2 - x_1} = \frac{7 - 2}{6 - 8} = \frac{5}{-2} = -\frac{5}{2} )

Since the second line is parallel to the first, it will have the same slope.

Now, we can use the slope and the given point (3, 8) to find the equation of the second line and then find another point on it. Using point-slope form:

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

Substituting the values, we get:

( y - 8 = -\frac{5}{2}(x - 3) )

Expanding and simplifying:

( y - 8 = -\frac{5}{2}x + \frac{15}{2} )

( y = -\frac{5}{2}x + \frac{31}{2} )

Now, we can choose any value for ( x ) and find the corresponding ( y ) to get another point on the line. Let's choose ( x = 0 ):

( y = -\frac{5}{2}(0) + \frac{31}{2} )

( y = \frac{31}{2} )

So, another point on the second line is (0, 15.5).

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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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