Given the relative vertices a (0, - 1) and C (2,5) of square ABCD, the coordinates of vertices B and D are obtained

Given the relative vertices a (0, - 1) and C (2,5) of square ABCD, the coordinates of vertices B and D are obtained

Let the midpoint of AC be m (x, y), then x = 0 + 22 = 1, y = − 1 + 52 = 2, m (x, y) = m (1, 2). Let the slope of AC be K, then k = 3, so the slope of BD is − 1K = − 13. So the equation of straight line BD is y − 2 = − 13 (x − 1) (1), and the equation of circle with m point as center and | Ma | as radius is

Given the square ABCD, point a (- 2, - 3), C (1,1), find the coordinates of vertex D

First, we can conclude that the coordinates of D are in the fourth quadrant
Because it is a square, so the spacing is the same, we can get the coordinates as (2,3)

It is known that ABCD is a square, a (- 1,2), C (3,6). Find the coordinates of the other two vertices B and D

Because point a (- 1,2), C (3,6)
Then the slope of the line AC k = (6-2) / (3 + 1) = 4 / 4 = 1
Then the angle between the straight line and x-axis and y-axis is 45 degrees
Because square ABCD
The angle between AB, ad, BC, CD and AC is 45 degrees
Then AB and CD are parallel to X axis, ad and BC are parallel to y axis
The line passing through point a parallel to the X axis is y = 2
The line passing through point C parallel to y axis is x = 3
Then point B (3,2)
The line passing through point a parallel to y axis is x = - 1
The line passing through point C parallel to X axis is y = 6
Then point d (- 1,6)