Step 1: Recall the distance formula: d = √[(x₂ - x₁)² + (y₂ - y₁)²]

Step 2: Identify the coordinates: Point A: (x₁, y₁) = (3, 4) Point B: (x₂, y₂) = (7, 1)

Step 3: Substitute into the formula: d = √[(7 - 3)² + (1 - 4)²] d = √[4² + (-3)²] d = √[16 + 9] d = √25 d = 5

Answer: The distance is 5 units

Use the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

$d = \sqrt{(7 - 3)^2 + (1 - 4)^2}$

$d = \sqrt{4^2 + (-3)^2}$

$d = \sqrt{16 + 9}$

$d = \sqrt{25} = 5$

Answer: 5 units

Step 1: Recall the midpoint formula: M = ((x₁ + x₂)/2, (y₁ + y₂)/2)

Step 2: Identify the coordinates: Point 1: (x₁, y₁) = (-2, 5) Point 2: (x₂, y₂) = (6, -3)

Step 3: Calculate x-coordinate of midpoint: x_m = (-2 + 6)/2 = 4/2 = 2

Step 4: Calculate y-coordinate of midpoint: y_m = (5 + (-3))/2 = 2/2 = 1

Step 5: Write the midpoint: M = (2, 1)

Answer: The midpoint is (2, 1)

Use the midpoint formula: $M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$

$M = \left(\frac{-2 + 6}{2}, \frac{5 + (-3)}{2}\right)$

$M = \left(\frac{4}{2}, \frac{2}{2}\right)$

$M = (2, 1)$

Answer: $(2, 1)$

Step 1: Recall the midpoint formula: M = ((x₁ + x₂)/2, (y₁ + y₂)/2)

Step 2: Set up equations using given information: M(4, 7) and A(2, 3) 4 = (2 + x_B)/2 7 = (3 + y_B)/2

Step 3: Solve for x-coordinate of B: 4 = (2 + x_B)/2 8 = 2 + x_B x_B = 6

Step 4: Solve for y-coordinate of B: 7 = (3 + y_B)/2 14 = 3 + y_B y_B = 11

Step 5: Verify: Midpoint = ((2 + 6)/2, (3 + 11)/2) = (8/2, 14/2) = (4, 7) ✓

Answer: Point B is at (6, 11)

Step 1: Understand the problem: The radius is the distance from center to any point on the circle So radius = distance from C to P

Step 2: Use the distance formula: d = √[(x₂ - x₁)² + (y₂ - y₁)²]

Step 3: Substitute C(2, -1) and P(5, 3): r = √[(5 - 2)² + (3 - (-1))²] r = √[3² + 4²] r = √[9 + 16] r = √25 r = 5

Step 4: Recognize the Pythagorean triple: This is a 3-4-5 right triangle

Answer: The radius is 5 units

Strategy: Show that the sides satisfy the Pythagorean Theorem.

Find all three side lengths:

$AB$: from $(0,0)$ to $(5,0)$ $AB = \sqrt{(5-0)^2 + (0-0)^2} = 5$

$BC$: from $(5,0)$ to $(5,12)$ $BC = \sqrt{(5-5)^2 + (12-0)^2} = 12$

$AC$: from $(0,0)$ to $(5,12)$ $AC = \sqrt{(5-0)^2 + (12-0)^2} = \sqrt{25 + 144} = \sqrt{169} = 13$

Check Pythagorean Theorem: $AB^2 + BC^2 = 5^2 + 12^2 = 25 + 144 = 169$ $AC^2 = 13^2 = 169$

Since $AB^2 + BC^2 = AC^2$, the triangle is a right triangle.

Answer: Yes, it's a right triangle (in fact, a 5-12-13 right triangle)

Step 1: Understand the constraints:

• Right angle at B
• C is on the x-axis, so y-coordinate = 0: C(x, 0)
• Need to find x

Step 2: Use perpendicular slopes: If angle at B is 90°, then AB ⊥ BC Slopes must multiply to -1

Step 3: Find slope of AB: m_AB = (8 - 2)/(5 - (-3)) m_AB = 6/8 = 3/4

Step 4: Find slope of BC: m_BC = (0 - 8)/(x - 5) m_BC = -8/(x - 5)

Step 5: Set up perpendicularity condition: m_AB × m_BC = -1 (3/4) × (-8/(x - 5)) = -1

Step 6: Solve for x: (3/4) × (-8)/(x - 5) = -1 -24/(4(x - 5)) = -1 -24 = -4(x - 5) -24 = -4x + 20 -44 = -4x x = 11

Step 7: Verify the perpendicularity: m_AB = 3/4 m_BC = -8/(11 - 5) = -8/6 = -4/3 Product: (3/4) × (-4/3) = -12/12 = -1 ✓

Answer: Point C is at (11, 0)