A line segment is bisected by a line with the equation # 5 y + 2 x = 1 #. If one end of the line segment is at #(3 ,4 )#, where is the other end?

1 Answer

the other is end is at #(-13/29, -134/29)#

Explanation:

The given line is #5y+2x=1#
The line perpendicular to this line and passing thru point #U(3, 4)# is
#y-4=(5/2)(x-3)# by the two-point form
or #5x-2y=7#
Simultaneous solution of
#2x+5y=1# and #5x-2y=7# yields point #I(37/29, -9/29)#

Let #v=#vertical distance from point #I(37/29, -9/29)# to #U(3, 4)#
#v=4--9/29=125/29#
Let #h=#horizontal distance from point #I(37/29, -9/29)# to #U(3, 4)#
#h=3-37/29=50/29#
Let the other end point be #D(x_o, y_o)#

#x_o=37/29-h=37/29-50/29= -13/29#
#y_o=-9/29-125/29=-134/29#

Check by distance formula from point #I(37/29, -9/29)# to #U(3, 4)#
#d_1=sqrt((3-37/29)^2+(4--9/29)^2)=(25sqrt(29))/29#

Check by distance formula from point #I(37/29, -9/29)# to #D(-13/29, -134/29)#
#d_2=sqrt((37/29--13/29)^2+(-9/29--134/29)^2)=(25sqrt(29))/29#

Therefore #d_1=d_2#

God bless....I hope the explanation is useful.