A line passes through #(9 ,2 )# and #( 3, 5 )#. A second line passes through #( 4, 1 )#. What is one other point that the second line may pass through if it is parallel to the first line?

Answer 1

#(0,3)#

For a second line to be parallel to the first, it must have the same slope, but not be the same line. In slope-intercept form, this means that the #b# portions of each #y=mx+b# equation must differ.
As we might recall, #m=(y_2-y_1)/(x_2-x_1)# Thus, for our first line:
#m_1 = (5-2)/(3-9) = 3/(-6) = -1/2#
We have #y_1 = m_1x_1+b#. Substituting (9, 2) for the x and y respectively...
#-> 2 = -1/2 (9)+b = -9/2 +b -> b = 13/2#

Giving us for the first line:

#y = -1/2 x + 13/2#.

NOTE: For the next part, we will be using prime notation, e.g. y', m'. These are not being used to denote Calculus derivatives, but rather to distinguish them from the original variable and slope.

For our second line, we must have the same slope, but a different intercept. We can find our Y intercept like we did above:

#y' = m'x'+b' -> 1 = -1/2 (4) + b = -2 + b -> b = 3 -> y' = -1/2 x + 3#
We know the y intercept is the value the function takes on when #x=0#. Thus, in this case:
#y'(0) = -1/2 (0) + 3 = 3#
Thus, one point this new parallel line could pass through is #(0,3)#.
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Answer 2

To find a point that the second line may pass through if it is parallel to the first line, we need to consider the slope of the first line. The slope of a line passing through two points ((x_1, y_1)) and ((x_2, y_2)) is given by:

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

Using the given points ((9, 2)) and ((3, 5)), we can calculate the slope of the first line. Then, since the second line is parallel to the first line, it will have the same slope.

Next, using the given point ((4, 1)) and the slope we calculated, we can use the point-slope form of a line to find another point on the second line.

The point-slope form of a line is:

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

where ( (x_1, y_1) ) is a point on the line and ( m ) is the slope.

By substituting the given point ((4, 1)) and the slope we calculated, we can find the equation of the second line. Then, by solving this equation for ( y ) when ( x ) takes a different value, we can find another point that the second line passes through.

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Answer from HIX Tutor

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.

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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