A line segment is bisected by a line with the equation # 4 y - 3 x = 2 #. If one end of the line segment is at #( 7 , 5 )#, where is the other end?

2 Answers
Jul 7, 2016

The other point is on the line #color(green)(4y-3x=5)#
(any point on this line will satisfy the given requirement)

Explanation:

Given
#color(white)("XXX")#The line #color(red)(4y-3x=2)#
and a point #color(purple)(""(7,5))#

Consider the vertical line through #color(purple)(""(7,5))#
This vertical line will intersect #color(red)(4y-3x=2)# at #color(red)(""(x=color(purple)(7),y))#
where #color(red)(y)# can be determined by solving
#color(white)("XXX")color(red)(4y-3xxcolor(purple)(7)=2)#

#color(white)("XXX")rarr color(red)(y=23/4)#

The distance from #color(purple)(""(7,5))# to #color(red)(""(7,23/4))#
is #color(red)(23/4)-color(purple)(5)=3/4#

So a point #color(green)(""(7,5+2xx3/4) = (7,13/2))# will be twice as far away from #color(purple)(""(7,5))# as the point #color(red)(""(7,23/4))# on the same vertical line

Note that any point on a line parallel to #color(red)(4y-3x=2# through #color(green)(""(7,13/2))# will also be twice as far away from #color(purple)(""(7,5))# as the point from #color(purple)(""(7,5))# on #color(red)(4y-3x=2)# to that point.
enter image source here

If #color(red)("L1")# parallel to #color(green)("L2")#
then #triangle ABC ~=triangle ADE#
#rarr abs(AB):abs(AD)=abs(AC):abs(AE)#

Since #color(red)(4y-3x=2)# has a slope of #color(red)(3/4)#

Our required line will also have a slope of #color(green)(3/4)#
and since it passes through #(""(7,13/2))#
using the slope-point form, we have
#color(white)("XXX")color(green)(y-13/2=3/4(x-7))#
or
#color(white)("XXX")color(green)4y-3x=5#
enter image source here

Jul 7, 2016

The other end-pt. lies on the line given by the eqn. #4y-3x=5.#

Explanation:

Suppose that the other end-pt. is #P(X,Y)#.

Let the given end-pt. be #Q(7,5),# and the given line be #L : 4y-3x=2.#

If #M# is the mid-pt. of the segment #PQ#, then co-ords. of #M# as obtained by using Section Formula for Mid-pt. are #=M((7+X)/2,(5+Y)/2)#

Now, #PQ# is bisected by #L# at #M#, so, #M in L.#

Therefore, co-ords. of #M# must satisfy the eqn. of #L.#

Hence, #4{(5+Y)/2}-3{(7+X)/2}=2 rArr 20+4Y-21-3X=4,# i.e., #4Y-3X=5,# as Sir Alan P. has readily derived!

This shows that :
(i) the co-ords. of other end-pt. can not be uniquely derived under the given conds.
(ii) What we can say about it (the other end-pt.) is that it lies on #: 4y-3x=5.# This eqn. represents a line #||# to #L#.