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?

Question

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?

Allison Allen · Accepted Answer

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

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.

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#

Evelyn Arboleda · Answer

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

Evelyn Arboleda · Answer

undefined To find the other end of the line segment bisected by the line $4y - 3x = 2$, given that one end is at (7, 5), we first need to find the midpoint of the line segment. Then, we can use the midpoint formula to find the coordinates of the other end.

First, let's find the midpoint of the line segment. We know that the midpoint is the average of the coordinates of the endpoints. So, if (x, y) is the midpoint, we have:

$$x_{\text{midpoint}} = \frac{x_1 + x_2}{2}$$
$$y_{\text{midpoint}} = \frac{y_1 + y_2}{2}$$

Given that one endpoint is at (7, 5), we can substitute these values into the midpoint formula:

$$x_{\text{midpoint}} = \frac{7 + x_2}{2}$$
$$y_{\text{midpoint}} = \frac{5 + y_2}{2}$$

Now, we need to find the coordinates of the other endpoint, which will be the solution to the system of equations formed by the line equation and the midpoint formula.

Substituting the equation of the line ($4y - 3x = 2$) into the midpoint formula, we get:

$$x_{\text{midpoint}} = \frac{7 + x_2}{2} \implies 3x_2 = 14 - x$$
$$y_{\text{midpoint}} = \frac{5 + y_2}{2} \implies 4y_2 = 10 - y$$

Now, we can solve this system of equations to find the values of $x$ and $y$, which represent the coordinates of the other end of the line segment.