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.!