1 3 Additional Practice Midpoint And Distance Answer Key

1, 3 Additional Practice: Midpoint and Distance Formula – Answers & Explanations

Finding the midpoint and distance between two points is a fundamental concept in coordinate geometry. Mastering these formulas is crucial for success in higher-level math courses. This comprehensive guide provides answers and detailed explanations for additional practice problems involving the midpoint and distance formulas, going beyond the typical examples found in textbooks. We'll cover a range of difficulty levels, helping you solidify your understanding and build confidence in tackling these problems.

Understanding the Formulas:

Before we dive into the problems, let's review the essential formulas:

• Midpoint Formula: The midpoint M of a line segment with endpoints (x₁, y₁) and (x₂, y₂) is given by: M = ( (x₁ + x₂)/2 , (y₁ + y₂)/2 )

• Distance Formula: The distance d between two points (x₁, y₁) and (x₂, y₂) is given by: d = √[(x₂ - x₁)² + (y₂ - y₁)²]

Additional Practice Problems & Solutions:

Let's tackle some additional practice problems, categorized for clarity.

Section 1: Basic Midpoint Problems

Problem 1.1: Find the midpoint of the line segment connecting the points A(-2, 5) and B(4, -1).

Solution 1.1:

Using the midpoint formula:

M = ( (-2 + 4)/2 , (5 + (-1))/2 ) = ( 2/2 , 4/2 ) = (1, 2)

Therefore, the midpoint is (1, 2).

Problem 1.2: The midpoint of a line segment is (3, -2). One endpoint is (1, 4). Find the coordinates of the other endpoint.

Solution 1.2:

Let the other endpoint be (x, y). Using the midpoint formula:

( (1 + x)/2 , (4 + y)/2 ) = (3, -2)

Equating the x-coordinates: (1 + x)/2 = 3 => 1 + x = 6 => x = 5

Equating the y-coordinates: (4 + y)/2 = -2 => 4 + y = -4 => y = -8

Therefore, the coordinates of the other endpoint are (5, -8).

Section 2: Basic Distance Problems

Problem 2.1: Find the distance between the points C(2, 3) and D(6, 7).

Solution 2.1:

Using the distance formula:

d = √[(6 - 2)² + (7 - 3)²] = √[4² + 4²] = √(16 + 16) = √32 = 4√2

Problem 2.2: Determine whether the triangle with vertices P(1, 2), Q(4, 6), and R(7, 4) is an isosceles triangle.

Solution 2.2:

An isosceles triangle has at least two sides of equal length. We need to calculate the distances between each pair of points:

• PQ = √[(4 - 1)² + (6 - 2)²] = √(3² + 4²) = √25 = 5
• QR = √[(7 - 4)² + (4 - 6)²] = √(3² + (-2)²) = √13
• RP = √[(1 - 7)² + (2 - 4)²] = √((-6)² + (-2)²) = √40 = 2√10

Since none of the distances are equal, the triangle is not an isosceles triangle.

Section 3: Intermediate Problems Combining Midpoint and Distance

Problem 3.1: The endpoints of a diameter of a circle are A(-1, 2) and B(7, -4). Find the coordinates of the center of the circle and the length of its radius.

Solution 3.1:

The center of the circle is the midpoint of the diameter. Using the midpoint formula:

Center = ( (-1 + 7)/2 , (2 + (-4))/2 ) = (3, -1)

The radius is half the length of the diameter. Using the distance formula to find the diameter's length:

Diameter = √[(7 - (-1))² + (-4 - 2)²] = √[8² + (-6)²] = √(64 + 36) = √100 = 10

Radius = Diameter/2 = 10/2 = 5

The center of the circle is (3, -1) and its radius is 5.

Problem 3.2: Point P(x, y) is equidistant from A(2, 1) and B(6, 5). Find an equation relating x and y.

Solution 3.2:

Since P is equidistant from A and B, the distance PA equals the distance PB:

PA = √[(x - 2)² + (y - 1)²] PB = √[(x - 6)² + (y - 5)²]

PA = PB => √[(x - 2)² + (y - 1)²] = √[(x - 6)² + (y - 5)²]

Squaring both sides: (x - 2)² + (y - 1)² = (x - 6)² + (y - 5)²

Expanding and simplifying: x² - 4x + 4 + y² - 2y + 1 = x² - 12x + 36 + y² - 10y + 25

-4x - 2y + 5 = -12x - 10y + 61

8x + 8y = 56

Dividing by 8: x + y = 7 This is the equation relating x and y.

Section 4: Advanced Problems

Problem 4.1: Find the area of the triangle with vertices R(-1, 1), S(3, 5), and T(5, -1).

Solution 4.1:

We can use the determinant method to find the area of the triangle:

Area = (1/2) |(-1)(5 - (-1)) + 3(-1 - 1) + 5(1 - 5)| = (1/2) |(-1)(6) + 3(-2) + 5(-4)| = (1/2) |-6 - 6 - 20| = (1/2) |-32| = 16 square units

Problem 4.2: Prove that the points A(1, 1), B(4, 7), and C(10, 11) are collinear.

Solution 4.2:

Points are collinear if the distance between any two points added to the distance between another two points equals the distance between the remaining two points. Let's calculate the distances:

AB = √[(4 - 1)² + (7 - 1)²] = √(3² + 6²) = √45 = 3√5 BC = √[(10 - 4)² + (11 - 7)²] = √(6² + 4²) = √52 = 2√13 AC = √[(10 - 1)² + (11 - 1)²] = √(9² + 10²) = √181

Since AB + BC ≠ AC, let's use the slope method. If the slopes between any two pairs of points are equal, the points are collinear.

Slope of AB = (7 - 1)/(4 - 1) = 6/3 = 2 Slope of BC = (11 - 7)/(10 - 4) = 4/6 = 2/3 Slope of AC = (11 - 1)/(10 - 1) = 10/9

The slopes are not equal, therefore, this approach doesn't prove collinearity. Let's use the area method. If the area of the triangle formed by these points is 0, the points are collinear. Using the determinant method:

Area = (1/2) |(1)(7 - 11) + 4(11 - 1) + 10(1 - 7)| = (1/2) |(1)(-4) + 4(10) + 10(-6)| = (1/2) |-4 + 40 - 60| = (1/2) |-24| = 12 Since the area isn't 0, the points are not collinear. There's a mistake in the problem statement; the points are not collinear.

Conclusion:

This comprehensive guide provides a thorough exploration of midpoint and distance formula problems, ranging from basic to advanced levels. Remember to practice regularly, focusing on understanding the underlying concepts rather than just memorizing formulas. By working through these examples and similar problems, you'll build a strong foundation in coordinate geometry and improve your problem-solving skills. The key to mastering these concepts is consistent practice and attention to detail in applying the formulas correctly. Don't hesitate to review the formulas and work through additional problems to solidify your understanding. Good luck!