simple coordinate geometry question?

Question

the opposite vertexs of square has coordinates (3,4) & (1,-1).find the coordinates of other vertexs.

Answer 1

Nothing in the question says that the sides of the square is parallel to the XY axis of the coordinate system.

So, instead of adding information that is not there by saying that it is a rectangle, treat it as a square but rotated at an angle to the X axis.

Introducing any other assumptions will lead to a wrong answer.

Labelling the square ABCD starting in the lower left going clockwise around the square, the point A=(1,-1) and C = (3,4)

If the origin is point O, the radial vector to A is OA and to C is OC.

Let the vector representing the diagonal be AC

As OA + AC = OC, then AC = OC - OA = C - A = (3,4) - (1,-1)

AC = (2, 5)

The midpoint of the diagonal is the center of the square is X, and AX = 1/2*AC

AX = (1,2.5)

Now the vector XB is exactly the same length as AX, but orthogonal. This means the dot product is zero.

So

XB.AX = 0 and ||XB|| = ||AX|| = sqrt(1+2.5*2.5) = 2.69

if the components of XB = (x,y) then

x*1 + y*2.5 = 0 => x = -2.5y

And given x*x + y*y = 1 + 2.5*2.5

2.5*2.5y*y + y*y = 1 + 2.5*2.5

y*y(1+2.5*2.5) = 1+2.5*2.5 => y*y = 1

Which means y = +1 or -1

and x = -2.5 or + 2.5

So XB = (-2.5, 1) and XD = (2.5,-1)

So all we need to do is find B and D or the vectors OB and OD

Using AB = AX + XB

AB = (1,2.5) + (-2.5,1) = (-1.5,3.5)

And AD = AX + XD

AD = (1,2.5) + (2.5,-1) = (3.5,1.5)

And also OA + AB = OB

So OB = (1,-1) + (-1.5,3.5) = (-0.5,2.5)

And OA +AD = OD

So OC = (1,-1) + (3.5, 1.5) = (4.5, 0.5)

So the answers are:

B = (-0.5,2.5)
C = (4.5,0.5)

Answer 2

For this you can find the vertexes by using the numbers of the points all ready given. One of the other vertexes will have the same x-coordinate as the first point and the same y-coordinate as the second given point, for the point (3,-1). The last vertex will have the same y-coordinate as the first point and the same x-coordinate as the second given point, for the point (1,4). Once again, the other to vertexes would be (3,-1) and (1,4).

Answer 3

The square is not aligned with the x and y axes but is on an angle.

The slope of the diagonal is (4+1)/(3-1) = 5/2.
The slope of the other diagonal is = -2/5.

The midpoint (x,y) = ((3+1)/2,(4-1)/2) = (2,3/2)
From the point (1,-1) to the midpoint (2,3/2)

(Δx,Δy) = (1,5/2)

To get from the midpoint to the other vertices move:

(Δx,Δy) = (-5/2,1) and (5/2,-1)

And you get

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

The other corners of the square are (-1/2, 5/2) and (9/2, 1/2).

Answer 4

You have many different ways to do the problem. Here is one of them.

Given square ABCD, A(3, 4), and C (1, -1), find the coordinates of B and D.

AC = √(2^2+5^2) = √29
AB = BC = CD = DA = AC/√2 = (√58)/2
Slope of AC = 5/2 = m1
Slope of BD = -2/5 = m2
Center of AC: (2, 3/2)
Equation of BD: y - 3/2 = -(2/5)(x - 2)
Solve for y, y = -0.4x+2.3

Let (x, -0.4x +1.9) be the vertex point.
(x-2)^2 + (-0.4x+2.3 -1.5)^2 = 29/4

Solve for x, x = -.5, 4.5

Plug in x, y = 2.5, .5

Therefore, the two points are; B(-.5, 2.5) and D(4.5, .5)

Answer 5

you in uncomplicated terms would desire to factors to define the equation of a line, so use the 1st 2 (properly-known) factors: slope, m = upward thrust / run = (22 - (-18)) / (20 - (-10)) = 4/3 the equation for a line in slope intercept style is y = mx + b. y = 4/3 x + b plug in a properly-known factor to locate the y intercept, b: b = y - 4/3 x = 22 - 4/3 *20 = -4.sixty seven so your equation is now y = 4/3 x - 4.sixty seven. you could examine this via utilising the 2d factor, plug in -10 for x, to work out in case you get -18 for y. y = 4/3 (-10) - 4.sixty seven = -18 interior the 0.33 set of stuff you're given the y fee, merely plug in 2 for y, and sparkling up for x. x = (y + 4.sixty seven)*(3/4) = (2 + 4.sixty seven) * (3/4) = 5

Answer 6

I don't see how you could make an actual "square" but if it is a rectangle, the other vertices would be (1, 4) and (3, -1).

Answer 7

I drew it to find out that this can't be a square:

If you notice, the rectagle you gave is 5 high and 2 wide. For this to be a square, the height and width has to be the same.

Answer 8

(3, -1) and (1,4)