A line passes through #(4 ,7 )# and #(2 ,8 )#. A second line passes through #(1 ,5 )#. What is one other point that the second line may pass through if it is parallel to the first line?

Question

A line passes through #(4    ,7  )# and #(2   ,8   )#.  A second line passes through #(1    ,5  )#.  What is one other point that the second line may pass through if it is parallel to the first line?

Maverick Smith · Accepted Answer

The point #(0, 11/2)#.

The equation of a line is #y=mx+q# The line that passes from #(4,7)# and #(2,8)# can be obtained substituting the coordinates as #x# and #y# and doing the system #7=4m+q# #8=2m+q# We can subtract the second from the first to remove #q# obtaining #7-8=4m-2m+\cancel(q)-\cancel(q)# #-1=2m# and #m=-1/2#. We can also calculate #q#, even if it is not relevant for our problem, using one of the two equations #7=-1/2*4+q# #7=-2+q# #q=9# The equation of the first line is then #y=-1/2x+9# The second line is parallel, it means that the slope is the same #-1/2#. So the equation of the parallel is #y=-1/2x+r# where #r# is the intercept of the second line. To find it we know that the line passes from #(1,5)#, then #5=-1/2*1+r# #5+1/2=r# #11/2=r# The equation of the parallel is #y=-1/2x+11/2#. To find another point in this line it is enough to give a value to #x# and calculate the #y#. For example #x=0#, #y=-1/2*0+11/2=11/2#, then the point #(0, 11/2)# is another point of the parallel.