A line segment is bisected by a line with the equation # 3 y - 7 x = 3 #. If one end of the line segment is at #(7 ,8 )#, where is the other end?

1 Answer
Jul 2, 2016

The other point is at #(7/29, 10 26/29)#

Explanation:

Write the equation of the given line in standard form:

#3y = 7x +3 rArr y = (7x)/3+1#

The gradient is #7/3#

We will have to assume that this line is the perpendicular bisector rather than just a bisector, else the question cannot be answered.

The line segment has a gradient of #-3/7#
and passes through the point #(7,8)#

Find its equation using: #y-y_1 = m(x-x_1)#
#y - 8 = -3/7(x-7)#

#y = (-3x)/7 +3+8 rArr y = (-3x)/7 +11#

The midpoint of the line segment is where the lines intersect. Solve them simultaneously by equating the y's.

#y = (7x)/3+1 and y = (-3x)/7 +11#
#y=y rArr (7x)/3+1 =(-3x)/7 +11#

#(7x)/3 + (3x)/7 = 10" "xx21#

#49x+9x = 210#

#58x = 210 rArr x =105/29# This gives y as #274/29#

This point is the midpoint of the line segment.

#(7 +x_2)/2 = 105/29 " and "(8+y_2)/2 = 274/29#

#x_2 = 210/29 -7 " and "y_2 = 548/29 -8#

The other point is at #(7/29, 10 26/29)#