Using Coordinates in Proofs

The radius of a circle is the distance between the center and any point on the circle. Since we know the center of Zosia's circle and one point contained in it we can find the radius by substituting the given coordinates into the Distance Formula and simplify.

We have been given the center of Tearrik's circle and a point on its circumference. The radius is the distance between these points. Let's substitute the given coordinates into the Distance Formula and simplify.

For Tiffaniqua's circle we know the endpoints of a diameter. We will first calculate the length of the diameter. Then, to find the radius, we will divide the result by 2.

The radius is half the length of the diameter. r=d/2=13/2

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 $E (\frac{a + b}{2}, \frac{c}{2})$	$D E = \frac{a + b}{2} - \frac{b}{2}$	$D E = \frac{1}{2} a$

	8 Theory slides
	13 Exercises - Grade E - A
	Each lesson is meant to take 1-2 classroom sessions

$d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$
Side	Endpoints	Substitute	Simplify
$\overline{S P}$	$S (- 2, - 3)$ and $P (- 5, 4)$	$S P =$ $(- 5 - (- 2))^{2} + (4 - (- 3))^{2}$	$S P = 58$
$\overline{P A}$	$P (- 5, 4)$ and $A (3, 5)$	$P A =$ $(3 - (- 5))^{2} + (5 - 4)^{2}$	$P A = 69$
$\overline{A T}$	$A (3, 5)$ and $T (6, - 2)$	$A T =$ $(6 - 3)^{2} + (- 2 - 5)^{2}$	$A T = 58$
$\overline{T S}$	$T (6, - 2)$ and $S (- 2, - 3)$	$T S =$ $(- 2 - 6)^{2} + (- 3 - (- 2))^{2}$	$T S = 69$

$M (\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2})$
Diagonal	Endpoints	Substitute	Simplify
$\overline{P T}$	$P (- 5, 4)$ and $T (6, - 2)$	$M_{1} (\frac{- 5 + 6}{2}, \frac{4 + ( - 2 )}{2})$	$M_{1} (\frac{1}{2}, 1)$
$\overline{S A}$	$S (- 2, - 3)$ and $A (3, 5)$	$M_{2} (\frac{- 2 + 3}{2}, \frac{- 3 + 5}{2})$	$M_{2} (\frac{1}{2}, 1)$

Sides	Product of Slopes
$\overline{A B}$ and $\overline{B C}$	$\frac{4}{3} (- \frac{3}{4}) = - 1 ✓$
$\overline{B C}$ and $\overline{D C}$	$- \frac{3}{4} (\frac{4}{3}) = - 1 ✓$
$\overline{D C}$ and $\overline{A D}$	$\frac{4}{3} (- \frac{3}{4}) = - 1 ✓$
$\overline{A D}$ and $\overline{A B}$	$- \frac{3}{4} (\frac{4}{3}) = - 1 ✓$

Side	Points	M(x_1+x_2/2,y_1+y_2/2)	Evaluate
PQ	( 1,4), ( - 3,4)	M(1+( -3)/2,4+ 4/2)	M(- 1,4)
QR	( - 3,4), ( - 3,- 2)	M(-3+( -3)/2,4+( -2)/2)	M(-3,1)

Quadrilateral	Definition
Parallelogram	Quadrilateral with two pairs of parallel sides
Rhombus	Parallelogram with four congruent sides
Rectangle	Parallelogram with four right angles
Square	Parallelogram with four congruent sides and four right angles

Side	Points	Slope Formula	Simplified
AB	( - 2,- 1), ( 4,2)	2-( - 1)/4-( - 2)	1/2
BC	( 4,2), ( 3, 4)	4- 2/3- 4	- 2
CD	( 3, 4), ( - 3, 1)	1- 4/- 3- 3	1/2
DA	( - 3, 1), ( - 2,- 1)	- 1- 1/- 2-( - 3)	- 2

Side	Points	Distance Formula	Simplified
AB	( - 2,- 1), ( 4,2)	sqrt(( 4-( - 2))^2+( 2-( - 1))^2)	3sqrt(5)
BC	( 4,2), ( 3, 4)	sqrt(( 3- 4)^2+( 4- 2)^2)	sqrt(5)
CD	( 3, 4), ( - 3, 1)	sqrt(( - 3- 3)^2+( 1- 4)^2)	3sqrt(5)
DA	( - 3, 1), ( - 2,- 1)	sqrt(( - 2-( - 3))^2+( - 1- 1)^2)	sqrt(5)

$m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}$
Side	Endpoints	Substitute	Simplify
$\overline{A B}$	$A (- 6, 2)$ and $B (- 3, 6)$	$m_{\overline{A B}} = \frac{6 - 2}{- 3 - ( - 6 )}$	$m_{\overline{A B}} = \frac{4}{3}$
$\overline{B C}$	$B (- 3, 6)$ and $C (9, - 3)$	$m_{\overline{B C}} = \frac{- 3 - 6}{9 - ( - 3 )}$	$m_{\overline{B C}} = - \frac{3}{4}$
$\overline{D C}$	$D (6, - 7)$ and $C (9, - 3)$	$m_{\overline{D C}} = \frac{- 3 - ( - 7 )}{9 - 6}$	$m_{\overline{D C}} = \frac{4}{3}$
$\overline{A D}$	$A (- 6, 2)$ and $D (6, - 7)$	$m_{\overline{A D}} = \frac{- 7 - 2}{6 - ( - 6 )}$	$m_{\overline{A D}} = - \frac{3}{4}$

Using Coordinates in Proofs

Catch-Up and Review

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

Hint

Solution

Proof

BC∥DE

DE=21​BC

Answer

Hint

Solution

Opposite Sides are Congruent

Diagonals Bisect Each Other

Proof

Proof

Hint

Solution

Finding Perpendicular Bisectors

Finding the Circumcenter

Finding Slopes

Finding Lengths

Finding the Length

Solving the Equation

Finding the Length

Solving the Equation

Using Coordinates in Proofs

Recommended exercises

$\overline{B C} ∥ \overline{D E}$

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