A line segment is bisected by line with the equation # 6 y + 7 x = 4 #. If one end of the line segment is at #(2 ,4 )#, where is the other end?

2 Answers
Oct 8, 2017

End point coordinates #((6/7),-6)#

Explanation:

It is assumed that the intersecting line is a perpendicular bisector.
#6y+7x=4#
#y=-(7/6)x+(4/6)=-(7/6)x+(2/3)#
Slope #= -(7/6)#
Slope of the line segment #=-1/-(7/6)=6/7#
Equation of line segment is
#(y-4)=(6/7)(x-2)#
#7y-28=6x-12#
#7x-6y=16#
Solving the equations we will get the coordinates of the mid point (intersection point).
#14x=20#
#x=(10/7)#
#10-6y=16#
#y=-1#
Mid point coordinates # ((10/7),-1)#
But mid point #(4+y1)/2=-1# & #(2+x1)/2=10/7#
#y1=-6# & #x1=(20/7)-2=6/7#
End point coordinates #((6/7),-6)#

Oct 8, 2017

#(-(18)/5,-4/5)#

Explanation:

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Let #L1# be the perpendicular bisector and #L2# (segment #AB#) be the bisected line, as shown in the figure.
Given that the equation of the bisector #L1# is #6y+7x=4#,
#=> y=-7/6x+2/3#
Let #m_1# be the slope of #L1#, and #m_2# the slope of #L2#,
#=> m_1=-7/6#
Recall that the product of the slopes of two perpendicular lines is #-1#,
#=> m_2xxm_1=-1, => m_2=6/7#
Given #A=(2,4)#,
#=># equation of #L2# is : #y-4=6/7(x-2)#
#=> y=6/7x+(16)/7#
Set the equations of #L1 and L2# equal to each other to find the intersection point #M(x_m, y_m)#, which is also the midpoint of #L2#.
#=> -7/6x+2/3=6/7x+16/7#
#=> x=-4/5#
#=> y=-7/6x+2/3=-7/6xx(-4/5)+2/3=8/5#
#=> M(x_m,y_m)=(-4/5,8/5)#
Let the other end point of #L2# be #B(x,y)#,
Since #M# is the midpoint of #L2#,
#=> ((x+2)/2, (y+4)/2) = (-4/5,8/5)#
#=> (x,y)=(-(18)/5,-4/5)#