Question 1123947


{{{ABCD}}} is a rectangle

The coordinates of {{{A}}} and {{{B}}} are ({{{1}}},{{{4}}}) and ({{{5}}},{{{2}}}) respectively. 

The x-coordinate of {{{D}}} is {{{-2}}}. -> {{{D}}} is at ({{{-2}}},{{{y}}})


Find:

a) the equation of {{{AD}}}

we will first find a slope of line that contais{{{AB}}}, a line that is perpendicular to the line that contains {{{AD}}} :


use given points ({{{1}}},{{{4}}}) and ({{{5}}},{{{2}}}) to find slope {{{m}}}

{{{m=(y[2]-y[1])/(x[2]-x[1])}}}

{{{m=(2-4)/(5-1)}}}

{{{m=-2/4}}

{{{m=-1/2}}

perpendicular lines have slopes that are negative reciprocal to each others, so the line{{{AD}}} will have a slope {{{m=-(1/(-1/2))}}->{{{m=2}}


so, so far we have {{{y= 2x+b}}}...use given point that lie on AB ({{{1}}},{{{4}}}) (

{{{4=2*1+b}}}
{{{4-2=b}}}
{{{b=2}}}

and, your equation o the line {{AD}}} is: {{{y=2x+2}}}


b) the y-coordinate of {{{D}}}

to find it, use {{{y=2)x+2}}} and ({{{-2}}},{{{y}}})
{{{y=2(-2)+2}}}
{{{y=-4+2}}}
{{{y=-2}}}
so,{{{D}}} is at  ({{{-2}}},{{{-2}}})


c) the coordinates of {{{C }}}

point {{{C }}} is intersection point o the line {{{DC}}} and {{{BC}}}

since the line {{{DC}}} is parallel to the line {{{AB}}}, they have same slope

{{{m=-1/2}}}

equation of the line {{{DC}}} is:

{{{y=-(1/2)x+b}}}....use point {{{D}}} which is at  ({{{-2}}},{{{-2}}})


{{{-2=-(1/2)(-2)+b}}}

{{{-2=1+b}}}

{{{-2-1=b}}}

{{{b=-3}}}

{{{y=-(1/2)x-3}}}.....eq.1

and equation of {{{AD}}} has a slope {{{m=2}}} and is parallel to {{{BC}}}

so,equation of the line {{{BC}}} is:
{{{y=2x+b}}}....use point {{{B}}} which is at  ({{{5}}},{{{2}}})

{{{2=2*5+b}}}

{{{2-10=b}}}

{{{b=-8}}}

and {{{y=2x-8}}}.....eq.2

from eq.1 and eq.2 we have

{{{2x-8=-(1/2)x-3}}}

{{{2x+(1/2)x=8-3}}}

{{{2x+x/2=5}}}

{{{4x+x=10}}}

{{{5x=10}}}

{{{x=2}}}
go to  {{{y=2x-8}}}.....eq.2, plug in {{{x}}}


 {{{y=2*2-8}}}
 {{{y=4-8}}}
 {{{y=-4}}}

so, point {{{C }}} is  at  ({{{2}}},{{{-4}}})


({{{1}}},{{{4}}}) and ({{{5}}},{{{2}}})

{{{drawing( 600, 600, -10, 10, -10, 10,
circle(1,4,.12),locate(1,4,A(1,4)),
circle(5,2,.12),locate(5,2,B(5,2)),
circle(-2,-2,.12),locate(-2,-2,D(-2,-2)),
circle(2,-4,.12),locate(2,-4,C(2,-4)),
green(line(1,4,5,2)),green(line(1,4,-2,-2)),
green(line(5,2,2,-4)),green(line(-2,-2,2,-4)),


 graph( 600, 600, -10, 10, -10, 10, 0)) }}}