Question 297859
Given 4 points, divide them into two sets. 
One set is the lengthwise component on the kite (line AB below). 
The other set is the transverse (sideways) component of the kite (line CD).
{{{drawing(300,300,-2,10,-2,10,
locate(1.5,6.25,C),
locate(6.25,2,D),
locate(0.25,0,A),
locate(6.25,6,B),
locate(4.5,4.2,I),
blue(line(0,0,6,6)),green(line(2,6,6,2)))}}} 
Calculate the eqaution of the line using the slope and the point-slope form of a line for the length wise component.
{{{y-ya=m1(x-xa)}}}
 {{{m1=(yb-ya)/(xb-xa)}}}
Then convert to slope-intercept form
{{{y=m1*x+b1}}}
Do the same for the transverse component to get its equation,
{{{y=m2*x+b2}}}
If it's a kite, the lines will be perpendicular and the relationship between the two slopes will be,
{{{m1*m2=-1}}}
If that's not the case, then stop there. It's not a kite.
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If it is true then find the intersection point, I, between the two lines. 
Solve the system of equations.
{{{y=m1*x+b1}}}
{{{y=m2*x+b2}}}
Since they both equal y, set them equal to each other.
{{{m1*xi+b1=m1*xi+b2}}}
Solve for x.
Then solve for y using either equation. 
Now you have the point of intersection (xi,yi) of the two components of the kite.
If it's a kite, the distance from the intersection point (xi,yi) to the ends of the transverse components (C and D) must be identical. 
{{{Dci=Ddi}}}
{{{Dci^2=Ddi^2}}}
{{{(xc-xi)^2+(yc-yi)^2=(xd-xi)^2+(yd-yi)^2}}}
If that's the case, then you've proven that the points form a kite. 
If it's not true, then you don't have a kite.