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

First off, even if it is crude, draw a sketch.

If the bisectors are perpendicular, then the products of their gradients will be #-1#

Let #L# be #2y+x=7#
#y=-1/2x+7/2#
I will be using #m# to denote the gradient (its easier than writing grad for formatting)
#:.m_L=-1/2#
Let #B=#bisector
#m_B*m_L=-1#
#:.m_B=2#

Now, we are searching for the equation of a line given a gradient and a point. This will allow us to easily find points on the bisector. We will use:

#y-y_1=m(x-x_1)# where #(x_1,y_1)# is a co-ordinate and #m# is the gradient.

#:.# eqn of B;

#y-3=2(x-5)#
#y-3=2x-10#
#y=2x-7#

Going back to our sketch, there is a point that these two lines cross. We can find this point by solving simultaneously for B and L

Subst B into L

#2(2x-7)+x=7#
#5x-14=7#
#5x=21#
#x=21/5#

#y=2x-7#
#y=2*21/5-7#
#y=7/5#

Let #M=(21/5, 7/5)#

We can now use column vectors to get us to Q. Since the part of the segment either side of the line bisecting it is equal, we know that the vector #vec(PM)=vec(MQ)#

#vec(PM)=((21/5-5),(7/5-3))#
#=((-4/5),(-8/5))#
#:.vec(MQ)=((-4/5),(-8/5))#
#Q=(21/5-4/5, 7/5-8/5)#
#Q=(17/5, -1/5)#

There is quite probably a simpler way to do this, but I don't know of any. Please tell me if I need to clarify anything