A line segment is bisected by a line with the equation # 7 y + x = 7 #. If one end of the line segment is at #(1 ,3 )#, where is the other end?

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Answer 1 · 2017-07-09T06:44:34

The other end is #=(0.4,-1.2)#

Let's rewrite the equation of the line.

#7y+x=7#

#7y=-x+7#

#y=-1/7x+1#................................#(1)#

The slope is #m=-1/7#

The slope of the perpendicular line is #m'=-1/m=-1/(-1/7)=7#

The equation of the perpendicular line is

#y-3=7(x-1)#

#y=7x-7+3=7x-4#.......................#(2)#

The point of intersection is obtained by solving for #x# and #y# in equations #(1)# and #(2)#

#7x-4=-1/7x+1#

#7x+1/7x=4+1=5#

#50/7x=5#

#x=5*7/50=7/10=0.7#

#y=7*0.7-4=4.9-0.4=0.9#

The point of intersection is #=(0.7,0.9)#

Let the other end of the line segment be #=(a,b)#

Then,

#((a+1)/2,(b+3)/2)=(0.7,0.9)#

#a+1=2*0.7=1.4#, #=># , #a=1.4-1=0.4#

#b+3=2*0.9=1.8#, #=>#, #b=1.8-3=-1.2#

The other end is #=(0.4,-1.2)#