A line segment is bisected by a line with the equation # - 6 y + 2 x = 4 #. If one end of the line segment is at #( 4 , 8 )#, where is the other end?

Question

A line segment is bisected by a line with the equation #  - 6  y  + 2  x   =   4  #.  If one end of the line segment is at #(    4  ,  8  )#, where is the other end?

Mason Adams · Accepted Answer

The other point is #B(42/5;-26/5)# the given line has slope #m=1/3# so the segment must have slope #m'=-3# and the line where lies the segment #AB# has equation #y=-3(x-4)+8#. If we find the instersection between these two lines, this represents the medium point between #A# and #B# By replacing the second equation into the first we obtain #2x-6(-3(x-4)+8)-4=0# that solved for #x# gives the coordinates of #M(31/5;7/5)# At this point if #B(t, u)#, it must be #(t+4)/2=31/5# and #(u+8)/2=7/5# that solved yeld #t=42/5# and #u=-26/5#

Allison Allen · Answer

undefined To find the other end of the line segment bisected by the line $ -6y + 2x = 4 $ with one end at (4, 8):

1. Find the slope of the given line. Rearrange the equation to slope-intercept form: $ y = \frac{1}{3}x - \frac{2}{3} $. The slope is $ \frac{1}{3} $.

2. Since the line bisects the line segment, the slope of the perpendicular bisector is the negative reciprocal of $ \frac{1}{3} $, which is $ -3 $.

3. Use the midpoint formula to find the midpoint of the line segment, which is also the point on the bisecting line: 
$$ \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) $$

4. Given one endpoint (4, 8), use the midpoint formula and the slope to find the other endpoint.

$$ x = \frac{4 + x_2}{2} $$
$$ y = \frac{8 + y_2}{2} $$

5. Substitute the equation of the bisecting line $ -6y + 2x = 4 $ into the equations above and solve for $ x $ and $ y $.

6. Once $ x $ and $ y $ are found, the coordinates $ (x, y) $ represent the other end of the line segment.