Equation of perpendicular bisector and coordinates of a point on a line?

Question

The point #C# lies on the perpendicular bisector of the line joining the points
#A (4, 6)# and #B (10, 2)#.
#C# also lies on the line parallel to #AB# through #(3, 11)#.
#(i)# Find the equation of the perpendicular bisector of #AB#. #[4]#
#(ii)# Calculate the coordinates of #C#. #[3]#

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Answer 1 · 2017-09-04T10:32:02

#" Eqn. of the p.b. of AB "(i) : 3x-2y-13=0.#

# (ii): C(9,7).#

#(i):#Let, #M# be the mid-point of the sgmt. #AB.# Denote, by #l_1,#

the line #AB,# and, #l_2# be its #bot#-bisector, with slope #m_2#.

#A=A(4,6), and, B=B(10,2) rArr M=M((4+10)/2),((6+2)/2))=M(7,4).#

Also, the slope #m_1# of #l_1=(6-2)/(4-10)=4/-6=-2/3.#

# l_1 bot l_2 rArr m_1*m_2=-1.#

# rArr m_2=-1/m_1=3/2.#

Thus, #M in l_2, , and, m_2=3/2;# hence, using the Slope-Point

Form for #l_2# we get, the following eqn. of the p.b. of #AB :#

#y-4=3/2(x-7), i.e., 2y-8=3x-21,#

# or, 3x-2y-13=0.....................................................(1).#

#(ii):# Let, #l'_1 || l_1# be the line through #C(3,11).#

#:." The slope of "l'_1="the slope of "l_1=-2/3, &, C in l'_1.#

#:." The eqn. of "l_1' : y-11=-2/3(x-3),#

#i.e., 3y-33=-2x+6, or, 2x+3y-39=0.............(2).#

#because, {C} in l'_1 nnl_2, :." solving "(1) and (2)," we get "C(9,7).#

Enjoy Maths.!