either prove or disprove the statement, the points (-5, -3), (-2, -4), and (5,1) are the vertices of a right triangle. use a graph only as a guide. let the three points (-5, -3), (-2, -4), and (5,1) be labeled as a, b, and c respectively. graph the triangle formed by the vertices a, b, and c. choose the correct graph below. a. graph b. graph c. graph d. graph by looking at the graph it seems that the points (-5, -3), (-2, -4), and (5,1) dropdown a right triangle.

Question

Accepted Answer

To determine if the points $ A(-5, -3) $, $ B(-2, -4) $, and $ C(5, 1) $ form a right triangle, we use the distance formula to find the lengths of the sides and then check the Pythagorean theorem.

### Step 1: Recall the distance formula
The distance between two points $ (x_1, y_1) $ and $ (x_2, y_2) $ is given by:
$$
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
$$

### Step 2: Calculate the length of $ AB $
For points $ A(-5, -3) $ and $ B(-2, -4) $:
$$
AB = \sqrt{(-2 - (-5))^2 + (-4 - (-3))^2} = \sqrt{(3)^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10}
$$

### Step 3: Calculate the length of $ BC $
For points $ B(-2, -4) $ and $ C(5, 1) $:
$$
BC = \sqrt{(5 - (-2))^2 + (1 - (-4))^2} = \sqrt{(7)^2 + (5)^2} = \sqrt{49 + 25} = \sqrt{74}
$$

### Step 4: Calculate the length of $ AC $
For points $ A(-5, -3) $ and $ C(5, 1) $:
$$
AC = \sqrt{(5 - (-5))^2 + (1 - (-3))^2} = \sqrt{(10)^2 + (4)^2} = \sqrt{100 + 16} = \sqrt{116}
$$

### Step 5: Check the Pythagorean theorem
We check if $ AB^2 + BC^2 = AC^2 $, $ AB^2 + AC^2 = BC^2 $, or $ BC^2 + AC^2 = AB^2 $.

- $ AB^2 = (\sqrt{10})^2 = 10 $
- $ BC^2 = (\sqrt{74})^2 = 74 $
- $ AC^2 = (\sqrt{116})^2 = 116 $

Now, $ AB^2 + BC^2 = 10 + 74 = 84 $, which is not equal to $ AC^2 = 116 $.

$ AB^2 + AC^2 = 10 + 116 = 126 $, which is not equal to $ BC^2 = 74 $.

$ BC^2 + AC^2 = 74 + 116 = 190 $, which is not equal to $ AB^2 = 10 $.

Wait, this suggests a mistake. Let's recalculate $ AC $:

For $ A(-5, -3) $ and $ C(5, 1) $:
$$
AC = \sqrt{(5 - (-5))^2 + (1 - (-3))^2} = \sqrt{(10)^2 + (4)^2} = \sqrt{100 + 16} = \sqrt{116} \approx 10.77
$$

Wait, maybe I made a mistake in the coordinates. Let's re-express the points:

$ A(-5, -3) $, $ B(-2, -4) $, $ C(5, 1) $

Let's recalculate the slopes to check for perpendicular lines (since a right triangle has two perpendicular sides, slopes are negative reciprocals).

Slope of $ AB $: $ m_{AB} = \frac{-4 - (-3)}{-2 - (-5)} = \frac{-1}{3} = -\frac{1}{3} $

Slope of $ BC $: $ m_{BC} = \frac{1 - (-4)}{5 - (-2)} = \frac{5}{7} $

Slope of $ AC $: $ m_{AC} = \frac{1 - (-3)}{5 - (-5)} = \frac{4}{10} = \frac{2}{5} $

Now, check if any two slopes are negative reciprocals (product = -1):

- $ m_{AB} \times m_{BC} = -\frac{1}{3} \times \frac{5}{7} = -\frac{5}{21}
eq -1 $
- $ m_{AB} \times m_{AC} = -\frac{1}{3} \times \frac{2}{5} = -\frac{2}{15}
eq -1 $
- $ m_{BC} \times m_{AC} = \frac{5}{7} \times \frac{2}{5} = \frac{2}{7}
eq -1 $

Wait, this contradicts the initial thought. But maybe the graph is misleading. Wait, let's recalculate the distances correctly.

Wait, $ AB $:

$ x_1 = -5, y_1 = -3 $; $ x_2 = -2, y_2 = -4 $

$ \Delta x = -2 - (-5) = 3 $; $ \Delta y = -4 - (-3) = -1 $

$ AB = \sqrt{3^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10} \approx 3.16 $

$ BC $:

$ x_1 = -2, y_1 = -4 $; $ x_2 = 5, y_2 = 1 $

$ \Delta x = 5 - (-2) = 7 $; $ \Delta y = 1 - (-4) = 5 $

$ BC = \sqrt{7^2 + 5^2} = \sqrt{49 + 25} = \sqrt{74} \approx 8.60 $

$ AC $:

$ x_1 = -5, y_1 = -3 $; $ x_2 = 5, y_2 = 1 $

$ \Delta x = 5 - (-5) = 10 $; $ \Delta y = 1 - (-3) = 4 $

$ AC = \sqrt{10^2 + 4^2} = \sqrt{100 + 16} = \sqrt{116} \approx 10.77 $

Now, check $ AB^2 + BC^2 = 10 + 74 = 84 $, $ AC^2 = 116 $. Not equal.

$ AB^2 + AC^2 = 10 + 116 = 126 $, $ BC^2 = 74 $. Not equal.

$ BC^2 + AC^2 = 74 + 116 = 190 $, $ AB^2 = 10 $. Not equal.

Wait, but maybe I labeled the points wrong. Let's check the graph. The correct graph is A, where A is (-5, -3), B is (-2, -4), and C is (5, 1). But according to the calculations, the sides do not satisfy the Pythagorean theorem. However, maybe I made a mistake in the slope or distance.

Wait, let's recalculate the slope of AB: $ (-4 - (-3))/(-2 - (-5)) = (-1)/3 = -1/3 $

Slope of AC: $ (1 - (-3))/(5 - (-5)) = 4/10 = 2/5 $

Slope of BC: $ (1 - (-4))/(5 - (-2)) = 5/7 $

None of the products of slopes are -1, so no right angle. Therefore, the points do not form a right triangle.

But wait, maybe the graph is different. Wait, the options are A, B, C, D. Let's check the coordinates again.

Wait, maybe the points are A(-5, -3), B(-2, -4), C(5, 1). Let's plot them:

- A: (-5, -3) is in the third quadrant.
- B: (-2, -4) is also in the third quadrant, closer to the y-axis.
- C: (5, 1) is in the first quadrant.

So the triangle has vertices in third, third, and first quadrants. The graph A shows this. But according to the calculations, it's not a right triangle.

But maybe I made a mistake. Let's check the distance between A and B, B and C, A and C again.

AB: from (-5, -3) to (-2, -4):

$ \Delta x = 3 $, $ \Delta y = -1 $, so distance $ \sqrt{9 + 1} = \sqrt{10} $

BC: from (-2, -4) to (5, 1):

$ \Delta x = 7 $, $ \Delta y = 5 $, distance $ \sqrt{49 + 25} = \sqrt{74} $

AC: from (-5, -3) to (5, 1):

$ \Delta x = 10 $, $ \Delta y = 4 $, distance $ \sqrt{100 + 16} = \sqrt{116} $

Now, check if $ (\sqrt{10})^2 + (\sqrt{74})^2 = (\sqrt{116})^2 $

$ 10 + 74 = 84 $, and $ 116
eq 84 $. So no.

Alternatively, maybe the right angle is at B? Let's check the vectors.

Vector BA: A - B = (-5 - (-2), -3 - (-4)) = (-3, 1)

Vector BC: C - B = (5 - (-2), 1 - (-4)) = (7, 5)

Dot product of BA and BC: (-3)(7) + (1)(5) = -21 + 5 = -16 ≠ 0

Vector AB: B - A = (3, -1)

Vector AC: C - A = (10, 4)

Dot product: (3)(10) + (-1)(4) = 30 - 4 = 26 ≠ 0

Vector CA: A - C = (-10, -4)

Vector CB: B - C = (-7, -5)

Dot product: (-10)(-7) + (-4)(-5) = 70 + 20 = 90 ≠ 0

So no dot product is zero, meaning no right angle. Therefore, the points do not form a right triangle.

But the question says "By looking at the graph it seems that the points... a right triangle." Maybe the graph is misleading, but according to calculations, it's not. Wait, maybe I made a mistake in the coordinates. Let me check again.

Wait, the points are (-5, -3), (-2, -4), and (5, 1). Let's recalculate the slopes:

Slope of AB: (-4 - (-3))/(-2 - (-5)) = (-1)/3 = -1/3

Slope of BC: (1 - (-4))/(5 - (-2)) = 5/7

Slope of AC: (1 - (-3))/(5 - (-5)) = 4/10 = 2/5

None of these slopes are negative reciprocals, so no right angle. Therefore, the answer is "do not form".

But wait, maybe the graph is different. Let me check the options. The correct graph is A, but according to calculations, it's not a right triangle. So the answer is "do not form".

# Explanation:
## Step1: Calculate distances using distance formula.
For $ AB $: $ \sqrt{(-2 - (-5))^2 + (-4 - (-3))^2} = \sqrt{10} $
For $ BC $: $ \sqrt{(5 - (-2))^2 + (1 - (-4))^2} = \sqrt{74} $
For $ AC $: $ \sqrt{(5 - (-5))^2 + (1 - (-3))^2} = \sqrt{116} $

## Step2: Check Pythagorean theorem.
$ (\sqrt{10})^2 + (\sqrt{74})^2 = 10 + 74 = 84
eq 116 = (\sqrt{116})^2 $
$ (\sqrt{10})^2 + (\sqrt{116})^2 = 10 + 116 = 126
eq 74 = (\sqrt{74})^2 $
$ (\sqrt{74})^2 + (\sqrt{116})^2 = 74 + 116 = 190
eq 10 = (\sqrt{10})^2 $

## Step3: Check slopes for perpendicularity.
Slopes: $ -\frac{1}{3} $, $ \frac{5}{7} $, $ \frac{2}{5} $. No negative reciprocals (product ≠ -1).

# Answer:
do not form

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QUESTION IMAGE

Question

Explanation:

Step 1: Recall the distance formula

Step 2: Calculate the length of \( AB \)

Step 3: Calculate the length of \( BC \)

Step 4: Calculate the length of \( AC \)

Step 5: Check the Pythagorean theorem

Step1: Calculate distances using distance formula.

Step2: Check Pythagorean theorem.

Step3: Check slopes for perpendicularity.

Answer: