A line segment is bisected by line with the equation # 3 y - 3 x = 1 #. If one end of the line segment is at #(2 ,5 )#, where is the other end?

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Answer 1 · 2018-05-12T02:48:10

#(14/3, 7/3)#

The perpendicular to

#-3x + 3y = 1 #

through #(2,5)# is

# 3x + 3y = 3(2)+3(5)=21#

Let's find where these meet. Subtracting equations,

#6x = 20#

#x = 10/3#

Adding equations,

#6y = 22#

#y = 11/3#

So one endpoint is # A(2,5)# and the bisection point is #B(10/3, 11/3). # The other endpoint #C# satisfies:

#C - B = B - A #

#C = 2B -A = (2(10/3) - 2, 2(11/3)-5) = (14/3, 7/3)#

Check:

The midpoint is

#B = 1/2 (A+C) = ( (2+14/3)/2, (5+7/3)/2) = (10/3, 11/3) quad sqrt #

Check the midpoint #B# is on #3y-3x=1#

#3( 11/3) -3 (10/3) = 1 quad sqrt#

Check perpendicularity. The y intercept is #D(0,1/3)#.

Let's check #(D-B) cdot(C -A) #

#(0 - 10/3, 1/3 - 11/3) cdot ( 14/3-2, 7/3-5) #

#= (0 - 10/3) ( 14/3-2) + ( 7/3-5)( 1/3 - 11/3) = 0 quad sqrt #

Zero dot product means perpendicularity.