Using Coordinates in Proofs

The incenter of a triangle is the point of intersection of the triangle's angle bisectors. Find the coordinates of the incenter of the equilateral triangle $A B C$ without using the Incenter Theorem. Answer in exact form.

A equilateral triangle with side length 6. Its angle bisectors have been drawn to the incenter.

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

Finding the x-coordinate

Finding the y-coordinate

Using Coordinates in Proofs

Recommended exercises

	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 $\overline{A B}$	$- \frac{3}{4} (\frac{4}{3}) = - 1 ✓$

$m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}$
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 the x-coordinate

Finding the y-coordinate

Using Coordinates in Proofs

Recommended exercises

$\overline{B C} ∥ \overline{D E}$

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