Question 1152746
Two vertices of a triangle are{{{ A}}}({{{3}}},{{{6}}}) and B({{{7}}},{{{12}}}).


a) Find the equation of line {{{AB}}}.

{{{y=mx+b}}}

use given points to find a slope:

{{{m=(y[2]-y[1])/(x[2]-x[1])=(12-6)/(7-3)=6/4=3/2}}}

now use point-slope formula to find equation:

{{{y-y[1]=m(x-x[1])}}}

{{{y-6=(3/2)(x-3)}}}

{{{y-6=(3/2)x-3(3/2)}}}

{{{y=(3/2)x-9/2+6}}}

{{{y=(3/2)x-9/2+12/2}}}

{{{y=(3/2)x+3/2}}}->the equation of line {{{AB}}}



b) Find the equation of the perpendicular bisector of line {{{AB}}}.

the equation of the perpendicular bisector of line {{{AB}}} will have a slope negative reciprocal which is

{{{m[p]=-1/m=-1/(3/2)=-2/3}}}

then,   use point-slope formula to find equation:

{{{y-y[1]=m[p](x-x[1])}}}

{{{y-6=-(2/3)(x-3)}}}

{{{y-6=-(2/3)x-3(-2/3)}}}

{{{y-6=-(2/3)x+6/3}}}

{{{y-6=-(2/3)x+2}}}

{{{y=-(2/3)x+2+6}}}

{{{y=-(2/3)x+8}}}
c) Given that AC is perpendicular to AB and the equation of line BC is {{{y=-5x+47}}},find the coordinates of C.
the coordinates of C will be the coordinates of intersection point, or solution of

{{{y=-(2/3)x+8}}}
{{{y=-5x+47}}}
---------------------------left sides are equal, then

{{{-(2/3)x+8=-5x+47}}}.............solve for {{{x}}}

{{{-(2/3)x+5x=-8+47}}}

{{{-2x/3+15x/3=39}}}

{{{13x/3=39}}}

{{{13x=39*3}}}

{{{x=(39/13)*3}}}

{{{x=3*3}}}

{{{x=9}}}

find {{{y}}}:

{{{y=-5*9+47}}}
{{{y=2}}}

intersection point is at: ({{{9}}},{{{2}}})



{{{drawing( 600, 600, -15, 15, -15, 15, 
circle(3,6,.12), locate(3.2,6,A(3,6)),
circle(7,12,.12), locate(7.2,12,B(7,12)),
circle(9,2,.12), locate(9.2,2,C(9,2)),
line(3,6,7,12),line(3,6,9,2),line(9,2,7,12),

 graph( 600, 600, -15, 15, -15, 15, -(2/3)x+8, -5x+47,(3/2)x+3/2)) }}}