A line segment is bisected by a line with the equation # 2 y - 2 x = 2 #. If one end of the line segment is at #( 3 , 8 )#, where is the other end?

Question

A line segment is bisected by a line with the equation #    2 y  - 2 x   =   2  #.  If one end of the line segment is at #(     3  ,  8  )#, where is the other end?

Jayden Jackson · Accepted Answer

The other end of the segment is #=(7,4)#

The line's equation is

#y-x=1#

#y=x+1#.......................#(1)#

The slope of the line is #m=1#

The slope of the segment is #m'#

#mm'=-1#

#m'=-1#

The segment's equation is

#y-8=-(x-3)#

#y=-x+3+8#

#y=-x+11#.......................#(2)#

Solving for #x# and #y# in the equations #(1)# and #(2)# gives the point of intersection of the line and the segment

#x+1=-x+11#

#2x=10#

#x=5#

#y=5+1=6#

The point of intersection is #=(5,6)#

Let the other end of the segment be #=(a,b)#

Consequently,

#(5,6)=((a+3)/2,(b+8)/2)#

#(a+3)/2=5#

#a=10-3=7#

#(b+8)/2=6#

#b=12-8=4#

Aurora Ayala · Answer

undefined Given the line equation $2y - 2x = 2$ can be simplified to $y - x = 1$, which bisects the line segment. If one end is at (3, 8), let the other end be at point $(x_2, y_2)$. The midpoint $(x_m, y_m)$ of the segment between $(3, 8)$ and $(x_2, y_2)$ lies on the line $y - x = 1$.

The midpoint formula gives us:
$$x_m = \frac{3 + x_2}{2}$$
$$y_m = \frac{8 + y_2}{2}$$

Since the midpoint lies on the line $y - x = 1$, we substitute $x_m$ and $y_m$ into the equation:
$$\frac{8 + y_2}{2} - \frac{3 + x_2}{2} = 1$$

Multiplying everything by 2 to clear the denominator:
$$8 + y_2 - 3 - x_2 = 2$$

Simplify:
$$y_2 - x_2 = 2 - 5$$
$$y_2 - x_2 = -3$$  --- (Equation 1)

Since the midpoint lies on the bisector, $x_m = y_m - 1$. Substituting the midpoint formula:
$$\frac{3 + x_2}{2} = \frac{8 + y_2}{2} - 1$$

Multiplying by 2 to clear the denominators:
$$3 + x_2 = 8 + y_2 - 2$$

Simplify:
$$x_2 + y_2 = 8 - 3 + 2$$
$$x_2 + y_2 = 7$$ --- (Equation 2)

Solving the system of Equations 1 and 2 ($y_2 - x_2 = -3$ and $x_2 + y_2 = 7$) simultaneously:

Adding the two equations:
$$2y_2 = 4$$
$$y_2 = 2$$

Substituting $y_2 = 2$ into $x_2 + y_2 = 7$:
$$x_2 + 2 = 7$$
$$x_2 = 5$$

Therefore, the other end of the line segment is at $(5, 2)$.