A line segment is bisected by a line with the equation # 4 y - 2 x = 5 #. If one end of the line segment is at #( 7 , 3 )#, where is the other end?

1 Answer
May 13, 2017

The other end is #(28/5,29/5)#

Explanation:

Given: #4y-2x=5" [1]"# is the equation of the bisector

Find the slope of the bisector:

#y = 2/4x+5/4#

#y = 1/2x+5/4#

The slope is #m_1= 1/2#

The of the bisected line is:

#m_2 = -1/m_1#

#m_2 = -1/(1/2)#

#m_2=-2#

Use the point-slope form of the equation of a line to find the equation of the bisected line:

#y = m(x-x_0)+y_0#

#y = -2(x-7)+3#

#y = -2x+14+3#

#y = -2x+17#

#y + 2x= 17" [2]"#

Add equation [2] to equation [1]:

#4y+y-2x+2x=5+17#

#5y = 22#

#y = 22/5#

Use equation [2] to find the corresponding value of x:

#22/5+2x=17#

#2x = 63/5#

#x = 63/10#

Use the midpoint formulas:

#x_"midpoint" = (x_"end"+x_"start")/2#

#y_"midpoint" = (y_"end"+x_"start")/2#

Substitute #63/10# for #x_"midpoint"# and 7 for #x_"start"#:

#63/10 = (x_"end"+7)/2#

Substitute #22/5# for #y_"midpoint"# and 3 for #y_"start"#:

#22/5 = (y_"end"+3)/2#

Solve for #x_"end" and y_"end"#

#63/10 = (x_"end"+7)/2#
#22/5 = (y_"end"+3)/2#

#x_"end" = 63/5-7#
#y_"end"= 44/5-3#

#x_"end" = 28/5#
#y_"end"= 29/5#

The other end is #(28/5,29/5)#