A line passes through #(8 ,9 )# and #(6 ,2 )#. A second line passes through #(3 ,4 )#. What is one other point that the second line may pass through if it is parallel to the first line?

1 Answer
reudhreghs
Apr 7, 2016

#(8,43/2)#

Explanation:

The gradient of the first line is given by #"rise"/"run"#, or the change in the #y# coordinate divided by the change in #x#. This is represented mathematically as

#(deltay)/(deltax) = (2-9)/(6-8) = 7/2#

which gives the gradient of the first line.

If two lines are parallel, then they have the same gradient, which means the gradient of the second line is also #7/2#.

Using the standard line equation, #y = mx + c#, where #m# is the gradient and #c# is any constant, we can plug in the values we do know from the point #(3,4)# to solve for #c# and find the full equation of the line.

#y = mx + c#
#4 = 7/2 * 3 + c#
#c = -13/2#
#y = 7/2x - 13/2#

Therefore, if we choose any random #x# value, say #x = 8#, and plug it in, we should get another point that is on the line.

#y = 7/2 * 8 - 13/2#
#y = 28 - 6.5 = 21.5 = 43/2#

It's always nicer to work in fractions, though sometimes it can be easier to convert to decimals. Mix and match where you feel one is easier than the other.

This gives the point

#(8, 43/2)#

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