Question 1179468

The vertices of a kite are 

A ({{{-3}}},{{{1}}}) 
B({{{0}}},{{{5}}})
C({{{4}}},{{{2}}}) 
D({{{2}}},{{{ -9}}})


Determine the lengths of the four sides of the kite. What do you notice?

the length of {{{AB}}} is:

{{{AB=sqrt((0-(-3))^2+(5-1)^2)}}}
{{{AB=sqrt(3^2+4^2)}}}
{{{AB=sqrt(25)}}}
{{{AB=5}}}

the length of {{{BC}}} is:

{{{BC=sqrt((4-0)^2+(2-5)^2)}}}
{{{BC=sqrt(4^2+(-3)^2)}}}
{{{BC=sqrt(25)}}}
{{{BC=5}}}

the length of {{{AD}}} is:

{{{AD=sqrt((2-(-3))^2+(-9-1)^2)}}}
{{{AD=sqrt(5^2+(-10)^2)}}}
{{{AD=sqrt(125)}}}
{{{AD=11.18}}}


the length of {{{CD}}} is:

{{{CD=sqrt((2-4)^2+(-9-2)^2)}}}
{{{CD=sqrt((-2)^2+(-11)^2)}}}
{{{CD=sqrt(125)}}}
{{{CD=11.18}}}

I  noticed that the length of {{{AB}}} is equal to the length of {{{BC}}}, and  the length of {{{AD}}} is equal to the length of {{{CD}}}


Determine the slopes of the diagonals (from one corner to the other corner)
What do you notice?

the diagonal {{{AC}}} lie on a line that contains vertices {{{A}}} and {{{C}}}

slope is {{{m=(y[2]-y[1])/(x[2]-x[1])}}}.......use A ({{{-3}}},{{{1}}}) and C({{{4}}},{{{2}}}) 

{{{m=(2-1)/(4-(-3))}}}

{{{m=1/(4+3)}}}
{{{m=1/7}}}

and  {{{BD}}} lie on a line that contains vertices B({{{0}}},{{{5}}}) and D({{{2}}},{{{ -9}}})

{{{m=(-9-5)/(2-0)}}}
{{{m=-14/2}}}
{{{m=-7}}}

=>I have noticed that slopes are negative reciprocal to each other which means the diagonals are perpendicular to each other


Find the midpoint of the diagonal AC.

A ({{{-3}}},{{{1}}}) and C({{{4}}},{{{2}}})

{{{M}}}=({{{(-3+4)/2}}},{{{(1+2)/2}}})
{{{M}}}=({{{1/2}}},{{{3/2}}})

Find the equation of the diagonal {{{BD}}}.

use  {{{m=-7}}} and point B({{{0}}},{{{5}}}), plug it in poit slope formula

{{{y-y[1]=m(x-x[1])}}}
{{{y-5=-7(x-0)}}}
{{{y-5=-7x}}}
{{{y=-7x+5}}}

Show the midpoint of AC lies on the equation of a line though BD.

the midpoint of AC is {{{M}}}=({{{1/2}}},{{{3/2}}})

{{{y=-7x+5}}}......substitute coordinates of the midpoint

{{{3/2=-7(1/2)+5}}}

{{{3/2=-7/2+10/2}}}

{{{3/2=3/2}}}-> same, so the midpoint of {{{AC}}} lies on the equation of a line though {{{BD}}}


{{{drawing ( 600, 600, -10, 10, -10, 10,
circle(-3,1,.12), locate(-3,1,A(-3,1)),
circle(0,5,.12), locate(0.3,5,B(0,5)),
circle(4,2,.12), locate(4,2,C(4,2)),
circle(2,-9,.12), locate(2,-9,D(2,-9)),
green(line(-3,1,0,5)),green(line(-3,1,2,-9)),
green(line(4,2,0,5)), green(line(4,2,2,-9)),
blue(line(0,5,2,-9)),blue(line(-3,1,4,2)),
circle(1/2,3/2,.12), locate(1/2,3/2,M(1/2,3/2)),

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