A line segment is bisected by a line with the equation # 9 y - 2 x = 5 #. If one end of the line segment is at #( 7 , 3 )#, where is the other end?

Question

A line segment is bisected by a line with the equation # 9  y  -  2  x   =   5  #.  If one end of the line segment is at #(    7  ,  3  )#, where is the other end?

Sophia Alfred · Accepted Answer

The general line containing all the endpoints of line segments with other endpoint #(7,3)# bisected by #9y-2x=5# is

#9y-2x=-3#

The endpoint of the perpendicular bisector is #(627/85, 111/85).# The other endpoint can be uniquely identified if the line is the segment's perpendicular bisector. If the line is just a bisector, as asked, each point on the line is the bisector of some segment whose endpoint is #(7,3)# and whose other endpoint sweeps out a curve. Let's see if we can figure out its equation. If we're told #(p,q)# is the midpoint of a segment with endpoint #(a,b)=(7,3)# the other endpoint #(x,y)# satisfies: # (x,y)-(p,q)=(p,q)-(a,b)# #(x,y)=(2p-a,2q-b)# #(x,y)=(2p-7,2q-3)# We have #(p,q)# on the line so #2p=9q-5#. #(x,y)=(9q-12,2q-3)=(-12,-3)+q(9,2)# The other endpoint #(x,y)# makes a line through #(-12,-3)# with direction vector #(9,2)# meaning slope #2/9.# Nonparametrically, that's the line #y+3 = 2/9(x+12)# #9y+27=2x+24# #9y-2x=-3# That's parallel to the original line, as far from it as #(7,3)# is. Now let's ask which #(x,y)# makes a perpendicular bisector to the line? The direction of the segment is #(x-a,y-b)=(x-7,y-3).# The direction of the line #-2x+9y=5# is #(9,2)# meaning for every #9# units #x# increases we increase #y# by #2# units. At zero dot product, we have perpendicularity: #(x-7,y-3) cdot(9,2) = 0# # 9(9q-12 - 7) + 2( 2q-3-3) = 0# #85q - 183 = 0# #q = 183/85# #(x,y) = (9q-12,2q-3)=(627/85, 111/85)# Check: # ( (627/85,111/85)-(7,3) ) cdot (9,2) = 0 quad sqrt#

Wyatt Anderson · Answer

undefined To find the other end of the line segment bisected by the line $9y - 2x = 5$, we can first find the midpoint of the segment using the given endpoint (7, 3). Then, we can use the midpoint to find the coordinates of the other end of the segment.

First, let's rearrange the equation of the line to $y = \frac{2x + 5}{9}$ to make it easier to work with. Then, we find the midpoint using the formula:

$$ \text{Midpoint} = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) $$

$$ \text{Midpoint} = \left( \frac{7 + x}{2}, \frac{3 + \frac{2x + 5}{9}}{2} \right) $$

Now, we'll equate the equation of the line with the y-coordinate of the midpoint to solve for $x$:

$$ \frac{2x + 5}{9} = \frac{3 + \frac{2x + 5}{9}}{2} $$

Solving this equation will give us the value of $x$, and then we can find the corresponding $y$-coordinate using the equation of the line. Once we have both $x$ and $y$, we'll have the coordinates of the other end of the line segment.