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

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

1 Answer
Ratnaker Mehta
Sep 4, 2017

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

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

Explanation:

(i):(i):Let, MM be the mid-point of the sgmt. AB.AB. Denote, by l_1,l1,

the line AB,AB, and, l_2l2 be its bot⊥-bisector, with slope m_2m2.

A=A(4,6), and, B=B(10,2) rArr M=M((4+10)/2),((6+2)/2))=M(7,4).A=A(4,6),and,B=B(10,2)⇒M=M(4+102),(6+22))=M(7,4).

Also, the slope m_1m1 of l_1=(6-2)/(4-10)=4/-6=-2/3.l1=6−24−10=4−6=−23.

l_1 bot l_2 rArr m_1*m_2=-1.l1⊥l2⇒m1⋅m2=−1.

rArr m_2=-1/m_1=3/2.⇒m2=−1m1=32.

Thus, M in l_2, , and, m_2=3/2;M∈l2,,and,m2=32; hence, using the Slope-Point

Form for l_2l2 we get, the following eqn. of the p.b. of AB :AB:

y-4=3/2(x-7), i.e., 2y-8=3x-21,y−4=32(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.!

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