Question

indicator competency benchmark competency benchmark + expressing geometric properties with equations □ calculate slope between points □ calculate distances between points using a² + b² = c² □ make claims about geometric shapes on a coordinate - plane using evidence 1a. find the slopes of ab, bc and ac using the slope formula. a (-3,1) b (2,2) c (-2,-4) 1b. find the lengths of ab, bc, and ac using the distance formula (which is a derivative of a² + b² = c²) 1c. classify this shape as strongly as possible and state why using the evidence above. you may want to use a² + b² = c².

Explanation:

Step1: Recall slope formula

The slope formula is $m=\frac{y_2 - y_1}{x_2 - x_1}$.
For line segment $AB$ with $A(-3,1)$ and $B(2,2)$:
$m_{AB}=\frac{2 - 1}{2-(-3)}=\frac{1}{5}$

Step2: Calculate slope of $BC$

For line segment $BC$ with $B(2,2)$ and $C(-2,-4)$:
$m_{BC}=\frac{-4 - 2}{-2 - 2}=\frac{-6}{-4}=\frac{3}{2}$

Step3: Calculate slope of $AC$

For line segment $AC$ with $A(-3,1)$ and $C(-2,-4)$:
$m_{AC}=\frac{-4 - 1}{-2-(-3)}=\frac{-5}{1}=- 5$

Step4: Recall distance formula

The distance formula is $d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$.
For $AB$:
$d_{AB}=\sqrt{(2 - (-3))^2+(2 - 1)^2}=\sqrt{25 + 1}=\sqrt{26}\approx5.099$

Step5: Calculate length of $BC$

For $BC$:
$d_{BC}=\sqrt{(-2 - 2)^2+(-4 - 2)^2}=\sqrt{16 + 36}=\sqrt{52}\approx7.211$

Step6: Calculate length of $AC$

For $AC$:
$d_{AC}=\sqrt{(-2-(-3))^2+(-4 - 1)^2}=\sqrt{1 + 25}=\sqrt{26}\approx5.099$

Step7: Classify the shape

Check the Pythagorean - theorem $a^{2}+b^{2}=c^{2}$.
$(d_{AB})^2+(d_{AC})^2=26 + 26=52=(d_{BC})^2$
Since the Pythagorean theorem holds, the shape is a right - triangle.

Answer:

1a. $m_{AB}=\frac{1}{5}$, $m_{BC}=\frac{3}{2}$, $m_{AC}=-5$
1b. $d_{AB}\approx5.099$, $d_{BC}\approx7.211$, $d_{AC}\approx5.099$
1c. Right - triangle because $(d_{AB})^2+(d_{AC})^2=(d_{BC})^2$