Using Coordinates in Proofs

Let's first plot the given vertices and the point of intersection of the diagonals.

By connecting these points, we can draw one side of the parallelogram and begin to draw the diagonals.

Let's recall the Parallelogram Diagonals Theorem.

Parallelogram Diagonals Theorem |- In a parallelogram, the diagonals bisect each other.

Therefore, we know that the missing parts of the diagonals will be congruent with AM and BM respectively. Let's draw the diagonals to their entire length and label the endpoints C and D.

To find the vertex opposite A we can use that the slope of both halves of the diagonal is the same.

We find the coordinates of C by moving two units to the right and one unit down from M(- 1,4). C(- 1 + 2,4 - 1) ⇔ C(1,3)

Like in Part A we will find the coordinates of the vertex by using that M is the midpoint of the diagonal.

We can now find the coordinates of D by subtracting 3 from both the x- and the y-coordinate of M(-1,4). D(- 1 - 3,4 - 3) ⇔ D(- 4,1)

Midpoint Formula: $(\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2})$
Road	Points	Substitute	Simplify
Road $1$	$B (4, 0)$ and $M (- 4, 4)$	$M_{1} (\frac{4 + ( - 4 )}{2}, \frac{0 + 4}{2})$	$M_{1} (0, 2)$
Road $2$	$M (- 4, 4)$ and $P (- 2, - 2)$	$M_{2} (\frac{- 4 + ( - 2 )}{2}, \frac{4 + ( - 2 )}{2})$	$M_{2} (- 3, 1)$

Distance Formula: $d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$
Points	Substitute	Simplify
$M_{1} (0, 2)$ and $M_{2} (- 3, 1)$	$d_{M_{1} M_{2}} = (- 3 - 0)^{2} + (1 - 2)^{2}$	$d_{M_{1} M_{2}} = 10$
$B (4, 0)$ and $P (- 2, - 2)$	$d_{B P} = (- 2 - 4)^{2} + (- 2 - 0)^{2}$	$d_{B P} = 210$

Slope Formula: $m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}$
Points	Substitute	Simplify
$M_{1} (0, 2)$ and $M_{2} (- 3, 1)$	$m_{M_{1} M_{2}} = \frac{1 - 2}{- 3 - 0}$	$m_{M_{1} M_{2}} = \frac{1}{3}$
$B (4, 0)$ and $P (- 2, - 2)$	$m_{B P} = \frac{- 2 - 0}{- 2 - 4}$	$m_{B P} = \frac{1}{3}$

$M (\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2})$
Segment	Endpoints	Substitute	Simplify
$\overline{B A}$	$B (0, 0)$ and $A (b, c)$	$D (\frac{0 + b}{2}, \frac{0 + c}{2})$	$D (\frac{b}{2}, \frac{c}{2})$
$\overline{C A}$	$C (a, 0)$ and $A (b, c)$	$E (\frac{a + b}{2}, \frac{0 + c}{2})$	$E (\frac{a + b}{2}, \frac{c}{2})$

Segment	Endpoints	Length	Simplify
$\overline{B C}$	$B (0, 0)$ and $C (a, 0)$	$B C = a - 0$	$B C = a$
$\overline{D E}$	$D (\frac{b}{2}, \frac{c}{2})$ and $E (\frac{a + b}{2}, \frac{c}{2})$	$D E = \frac{a + b}{2} - \frac{b}{2}$	$D E = \frac{1}{2} a$

Using Coordinates in Proofs

Catch-Up and Review

Investigating a Parallelogram in a Coordinate Plane

Proving Statements Using Different Methods

Hint

Solution

Ramsha's Solution

Finding the Equation of the Circle

Verifying If the Coordinates of the Point Satisfy the Equation

Dylan's Solution

Finding the Radius of the Circle

Calculating the Distance From the Point to the Center of the Circle

Solving Real Life Problems Involving Triangles

Hint

Solution

Triangle Midsegment Theorem

Proof

$\overline{B C} ∥ \overline{D E}$

$D E = \frac{1}{2} B C$

Solving Real Life Problems Involving Rectangles

Answer

Hint

Solution

Opposite Sides are Congruent

Diagonals Bisect Each Other

Properties of Parallelograms

Parallelogram Opposite Sides Theorem

Proof

Parallelogram Diagonals Theorem

Proof

Classifying a Parallelogram in a Coordinate Plane

Hint

Solution