Name the location of a fourth point
t that would form a square when the point are connected by line segments

i believe it is 2

lolll

answer is  

a.

7 in. x 11 in.

hope you get this right.

step-by-step explanation:

the line y=x is the line where the x and y coordinates of each point, are equal.

so this line is the line through the origin (0, 0), which makes an angle of 45° with the x-axis.

we draw triangle abc. to reflect this triangle with respect to the line y=x, we draw perpendiculars from the points a, b and c to to line y=x, and extend the perpendiculars so that the lint y=x is a perpendicular bisector of these line segments. 

we see that a(5, 3) is mapped to a'(3, 5)

and all the other points p(a, b) of the original triangle are mapped to p'(b,a)

the last statement is true for each point x and its reflection x', also it is true that the line y=x is the perpendicular bisector of xx' and yy'.

to prove these, check the second picture.

let the coordinates of x be (a, b), we want to find the coordinates of its reflection point with respect to the line y=x, point x'.

let m be the point on the line y=x, such that xm=mx'

let n be the point (b, b) which surely is on the line.

now, x'm=mx, nm is common and m(x'mn)=m(xmn)=90°, so by side-angle-side congruence postulate, the triangles x'mn and xmn

  are congruent, so nx=nx'

clearly the x coordinate of x' is b, so the y coordinate of x' is 

b+nx'=b+nx=a

so the coordinates of x' are (b, a)

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