Quadrilateral with given vertices

Question

If ABCD is a quadrilateral with vertices A(-3,0), B(-1,-3), C(2,-1) and D(0,2),

i) How can I determine if the coordinates of M is the midpoint of AC,

ii) How can I show that AC and BD bisect each other,

iii) How can I prove that angle ADC = 90°

iv) and how can I show that ABCD is a square?

Answer 1

Start with iii). Calculate the slope of both AD and DC. If one is the negative reciprocal of the other, then the lines are perpendicular. If angle ADC is not a right angle, then the figure is not a square. Do the same for the other 2 sides.

iv) Use the distance formula to calculate the length of each of the sides. All four sides of a square must be equal. If the angles at B and D are both right angles and the sides of the figure are equal, plane geometry tells you that the figure is a square.

i) and ii) If the figure is a square, then plane geometry tells you that the diagonals are equal and bisect eqch other.

If the figure is not a square, but a rectangle, the diagonals are still equal and bisect each other.

If (a,b) and (c,d) are the co-ordinates of the endpoints of a line segment, then the co-ordinates of the mid point are [(a+c)/2,(b+d)/2].

Answer 2

i) Consider what the vector from A to C will be, where it's going to be some vector v such that A + v = C. What's half of that vector? The midpoint will be point A plus half of the vector.

ii) In a similar fashion to i), you want to find the midpoints of AC and BD. You should be able to see that they have the same midpoint, and therefore they bisect each other.

iii) You can consider the angle ∠ADC as two vectors DA and DC. In order to find the angle, you want to use the formula cos(θ) = (u ⋅ v) / (|| u || || v ||), where u, v are the vectors DA and DC.

iv) In order to demonstrate a quadrilateral is a square, you need to be able to demonstrate that all sides are congruent and that all angles are 90 degrees. Simply show that all the sides are congruent by using the distance formula, and determine the angles using the formula aforementioned.