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

Question

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

Mason Adams · Accepted Answer

(151/13,27/13)

Rewrite the equation of the bisector in the form #y=mx+b#
#y=2/3x - 2/3#
Therefore its gradient equals to #2/3#.
Since the gradient product of two perpendicular lines is #-1#, the slope of the unknown line segment should be #(-1)/(2/3)=-3/2#

Solve for the equation of the line segment (which includes the point #(7,9)#. #y=-3/2(x-7)+9#, simplify (optional) and you get #y=-3/2x+39/2#

Solve for the #x#-coordinate for the intersection of the two perpendicular lines: Let #-3/2(x-7)+9=2/3x - 2/3 ", " x=121/13#

Now you can find the #x#-coordinate of the other end of the segment. Using the formula #x_1-x=-(x_2-x)#, where #x# is the #x#-coordinate of the point in the middle. #x_1=7# and #x_2=2*x-x_1#, #x_2=151/13#

Solve for the #y#-coordinate using the equation for the line segment. #y_2=-3/2(x-7)+9=27/13#
Therefore coordinate for the other end point is #(151/13,27/13)#

Wyatt Anderson · Answer

undefined To find the coordinates of the other end of the line segment, we need to determine the point where the given line bisects it. First, we need to find the slope of the given line by rearranging the equation into slope-intercept form, which is $y = mx + b$, where $m$ is the slope and $b$ is the y-intercept.

Given equation: $-3y + 2x = 2$

Rearranging the equation:

$-3y = -2x + 2$

$y = \frac{-2x + 2}{-3}$

$y = \frac{2}{3}x - \frac{2}{3}$

So, the slope of the given line is $m = \frac{2}{3}$.

Since the line bisects the segment, its slope should be perpendicular to the line segment. Therefore, the slope of the line bisecting the segment is the negative reciprocal of $\frac{2}{3}$, which is $-\frac{3}{2}$.

Now, we use the midpoint formula to find the coordinates of the other end of the line segment. Let $(x_1, y_1)$ be the given endpoint, which is $(7, 9)$, and $(x_2, y_2)$ be the coordinates of the other end.

Midpoint formula:

$$x_2 = \frac{x_1 + x_2}{2}$$
$$y_2 = \frac{y_1 + y_2}{2}$$

Substituting the given endpoint and the slope of the line bisecting the segment into the midpoint formula, we get:

$$x_2 = \frac{7 + x_2}{2}$$
$$y_2 = \frac{9 + y_2}{2}$$

Solving these equations simultaneously, we find the coordinates of the other end of the line segment:

$$x_2 = 7$$
$$y_2 = 6$$

So, the other end of the line segment is at $(7, 6)$.