A line segment is bisected by a line with the equation # 4 y + x = 3 #. If one end of the line segment is at #(5 ,6 )#, where is the other end?

1 Answer
Feb 25, 2018

The other end is at #(33/17,-106/17)#

Explanation:

Given the equation of the perpendicular bisector of the form:

#Ax+By = C_1#

The equation of the bisected line is of the form:

#Bx-Ay=C_2#

Writing the equation of the given perpendicular bisector in the above form:

#x+4y = 3#

Now, we can with the general form for the bisected line:

#4x-y = C_2#

To find the value of #C_2#, we substitute the point #(5,6)#:

#4(5)-6 = C_2#

#C_2 = 14#

The equation of the bisected line is:

#4x-y = 14#

Find the point of intersection of the lines:

#x+4y = 3#
#4x-y = 14#

Multiply the second equation by 4 and add to the first equation:

#17x= 59#

Solve for x:

#x = 59/17#

Use the given line to solve for y:

#59/17+4y = 3#

#y = -2/17#

We can use the midpoint equations to find the other end:

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

#y_"mid" = (y_"start"+y_"end")/2#

Substitute #(5,6)# for the starts and #(59/17,-2/17)# for mids:

#59/17 = (5+x_"end")/2#

#-2/17 = (6+y_"end")/2#

Solve for #(x_"end", y_"end")#

#x_"end" = 33/17#

#y_"end" = -106/17#

The other end is at #(33/17,-106/17)#