A line passes through #(4 ,9 )# and #(7 ,4 )#. A second line passes through #(8 ,7 )#. What is one other point that the second line may pass through if it is parallel to the first line?

Answer 1

#y=-5/3x+61/3#

As the two lines are parallel they both have the same gradient.

gradient (slope) #->("change in the y-axis")/("change in the x-axis")#
Let point 1 be #P_1->(x_1,y_1)=(4,9)# Let point 2 be #P_2->(x_2,y_2)=(7,4)#
Let point 3 be #P_3->(x_3,y_3) =(8,7)#
For the first line reading left to right #->x_1 to x_2#
So gradient is #P_2-P_1 ->(y_2-y_1)/(x_2-x_1) = (4-9)/(7-4) =-5/3#
Thus the gradient is # m=-5/3#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Thus equation for the second line #-> y=-5/3x+c#
This passes through the point #P_3 ->(x_3,y_3) =(8,7)#
#=> y_3=-5/3x_3+c" "->" "7=-5/3(8)+c#
#=>c=7+5/3(8) = 61/3#

Thus the equation of the second line is:

#y=-5/3x+61/3#
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Answer 2

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

The slope of the first line can be found using the formula:

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

Substituting the given points for the first line:

[ m = \frac{4 - 9}{7 - 4} ]

[ m = \frac{-5}{3} ]

Since the second line is parallel, it will also have a slope of ( -\frac{5}{3} ).

Now, we can find the equation of the second line using the point-slope form:

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

Substituting the given point ((8, 7)) and the slope ( -\frac{5}{3} ):

[ y - 7 = -\frac{5}{3}(x - 8) ]

Solving for (y):

[ y - 7 = -\frac{5}{3}x + \frac{40}{3} ]

[ y = -\frac{5}{3}x + \frac{40}{3} + 7 ]

[ y = -\frac{5}{3}x + \frac{40}{3} + \frac{21}{3} ]

[ y = -\frac{5}{3}x + \frac{61}{3} ]

This equation represents the second line. To find another point on this line, we can choose any arbitrary value for (x) and solve for (y). Let's choose (x = 5):

[ y = -\frac{5}{3}(5) + \frac{61}{3} ]

[ y = -\frac{25}{3} + \frac{61}{3} ]

[ y = \frac{36}{3} ]

[ y = 12 ]

So, another point that the second line may pass through is ((5, 12)).

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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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