Question

The two opposite vertices of a square are at (3,4) and (1,-1). What will be the other two vertices of the square?

Answer 1

Being totally reworked _ I had misunderstood the question.

`P_2->(x_2,y_2)=(9/2,1/2)`
`P_4->(x_4,y_4)=(-1/2,5/2)`

Explanation:

Set point 1 `->P_1->(x_1,y_1)=(3,4)` as given.

Set point 2 `->P_2->(x_2,y_2)` to be determined

Set point 3 `->P_3->(x_3,y_3)=(1,-1)` as given.

Set point 4 `->P_4->(x_4,y_4)` to be determined

Set centre point `P_c->(x_c,y_c)`

Note that the centre is calculated to be mid point between `P_1 and P_3->P_c->(x_c,y_c)=(2,3/2)`
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If there is any doubt it is always worth drawing a very quick sketch.

`color(brown)("Diagram1: This is as per the actual situation with precalculated points")`

`color(brown)("Diagram 2: Suppose we move the square such that it is central so that we")` `color(brown)("can see what is actually happening when rotated.")`

`color(brown)("In diagram 2 notice that the x and y values have swapped round when you compare")` `color(brown)("Point 1 to Point 4")`

`color(brown)("Moving back the Diagram 1 of actual")`

`color(blue)("Determine "P_4" - using "P_1" as the reference.")`
`y_4-y_c=x_1-x_c`
`y_4-3/2=3-2`
`color(red)(y_4=(3-2)+3/2=5/2)`

`x_c-x_4=y_1-y_c`
`2-x_4=4-3/2`

`color(red)(x_4=2-4+3/2= -1/2)`

`color(red)(P_4->(x_4,y_4)=(-1/2,5/2))`
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`color(blue)("Determine "P_2" - using "P_3" as the reference.")`

Watch the signs!!!!

`y_c-y_2=x_c-x_3`

`2-y_2= 3/2-(-1)`
`color(red)(y_2=3/2+1-2 = +1/2 )`

`x_2-x_c=y_c-y_3`
`x_2-2=3/2-(-1)`
`color(red)(x_2=3/2+1+2 = +9/2)`

`color(red)(P_2->(x_2,y_2)=(9/2,1/2))`

Answer 2

`(9/2, 1/2) and (-1/2, 5/2)`

Explanation:

Fun. The center of the square is the midpoint of the diagonal,

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

The difference between a corner and the center is called the direction vector from the center to the corner:

`(3,4)-(2,3/2)=(1,5/2)`

From the center we use the direction vector to get back to our two endpoints:

`(3,4) = (2,3/2) + (1,5/2)`

`(1,-1) = (2, 3/2) - (1,5/2)`

The perpendicular direction vector comes from swapping the coordinates and negating one. So the other corners of our square are:

`(2,3/2) + (5/2, -1) = (9/2, 1/2)`

`(2, 3/2) - (5/2, -1) = (-1/2, 5/2)`

Clearly the midpoint of the new diagonal is the same as the old, if we've done our arithmetic correctly. The diagonals should be perpendicular as well, so we're good.

If direction vectors are new to you, you can think of them as slopes with magnitudes. Our original direction vector of `(x,y)=(1,5/2)` is a slope of `y/x = 5/2`. When we add it to a point, we move along that slope by a distance `sqrt{x^2+y^2)= sqrt{29}/2` if I've done that right. It should be half the diagonal, `sqrt{(3-1)^2+(4--1)^2}=sqrt{29}.`