A line passes through #(9 ,2 )# and #( 4, 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

#(-6,7)#

Two parallel lines have the same gradient.

When we calculate the gradient of the first line,

gradient = #(y_2-y_1)/(x_2-x_1)#
#(5-2)/(4-9)=-3/5#
The gradient of the second line will also be #-3/5# We already have the coordinates #(4,1)# for the second line,
Now let's say that it passes through the point #(x_2,y_2).# We can take #y_2= 7.#
This is to simplify later, we can take any number for #y_2# because there are infinitely many points on the line.
So gradient = #-3/5# = #(7-1)/(x_2-4)#
#5(7-1) = -3(x_2-4)#
#30=-3x_2+12#
#30-12=-3x_2#
#x_2=18/-3#
#x_2 =-6#
So the coordinates of a second point are #(-6,7)#
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