Distance Formula
The Distance Formula is a crucial concept in coordinate geometry, enabling us to calculate the distance between two points in a two-dimensional or three-dimensional space. Given two points P(x1, y1) and Q(x2, y2), the distance d between them is calculated as:
d = sqrt((x2 - x1)2 + (y2 - y1)2)
Derivation of the Distance Formula
The Distance Formula can be derived from the Pythagorean Theorem, which states that in a right-angled triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. If we consider two points P and Q on a coordinate plane, we can form a right-angled triangle with PQ as the hypotenuse. The horizontal and vertical sides of the triangle represent the differences in the x-coordinates and y-coordinates of P and Q, respectively. Thus, the Distance Formula emerges as a direct application of the Pythagorean Theorem.
Key Points
- Square Root: The square root ensures that the distance is non-negative.
- Squaring Differences: Squaring the differences of the coordinates eliminates any negative values that might arise from subtracting coordinates.
- Summation: The sum of the squared differences of the x-coordinates and y-coordinates is taken under the square root.
Example Question 1
Find the distance between the points A(3, 4) and B(7, 1).
Solution: Using the Distance Formula: d = sqrt((7 - 3)2 + (1 - 4)2) d = sqrt((4)2 + (-3)2) d = sqrt(16 + 9) d = sqrt(25) d = 5
Thus, the distance between points A and B is 5 units.
Midpoint Formula
The Midpoint Formula is used to find the midpoint of a line segment connecting two points in a coordinate plane. Given two points P(x1, y1) and Q(x2, y2), the midpoint M is calculated as:
M = ((x1 + x2)/2, (y1 + y2)/2)
Derivation of the Midpoint Formula
The Midpoint Formula is derived by taking the average of the x-coordinates and y-coordinates of the two endpoints of a line segment. The midpoint is essentially the arithmetic mean of the x-coordinates and y-coordinates, and it represents a point that is equidistant from both endpoints of the line segment.
Key Points
- Averaging x-coordinates: The x-coordinate of the midpoint is the average of the x-coordinates of the two points.
- Averaging y-coordinates: Similarly, the y-coordinate of the midpoint is the average of the y-coordinates of the two points.
Example Question 2
Find the midpoint of the line segment connecting C(2, 3) and D(8, 5).
Solution: Using the Midpoint Formula: M = ((2 + 8)/2, (3 + 5)/2) M = (10/2, 8/2) M = (5, 4)
Therefore, the midpoint of the line segment CD is (5, 4).
Applications in Geometry
Perpendicular Bisector
A perpendicular bisector of a line segment is a line that is perpendicular to the segment and bisects it into two equal parts. The midpoint formula can be used to find the point where the segment is bisected, and the slope of the perpendicular bisector can be found using the negative reciprocal of the original line segment's slope. This concept is closely related to understanding parallel and perpendicular lines in coordinate geometry.
Example Question 3
Find the equation of the perpendicular bisector of the line segment connecting E(1, 2) and F(7, 6).
Solution:
1. Find the Midpoint: M = ((1 + 7)/2, (2 + 6)/2) M = (4, 4)
2. Find the Slope of EF: mEF = (y2 - y1)/(x2 - x1) mEF = (6 - 2)/(7 - 1) mEF = 4/6 mEF = ⅔
3. Find the Slope of the Perpendicular Bisector: mPB = -1/mEF mPB = -3/2
4. Find the Equation of the Perpendicular Bisector: Using the point-slope form of a line equation y - y1 = m(x - x1), where M(4, 4) and mPB = -3/2: y - 4 = -3/2(x - 4) y - 4 = -3/2x + 6 y = -3/2x + 10
Thus, the equation of the perpendicular bisector is y = -3/2x + 10. For further exploration of slopes and their applications in geometry, consider reviewing the basics of slope.
Understanding the properties of 3D shapes and their surface area and volume can also enhance the application of coordinate geometry in three-dimensional spaces, providing a deeper insight into geometric concepts.
FAQ
The Midpoint Formula finds extensive applications in various real-world scenarios, particularly in fields like physics, engineering, and computer graphics. For instance, in computer graphics, it is used to find the centre point of a line segment to perform operations like rotation, scaling, and translation. In physics, it can be used to determine the centre of mass of a two-particle system given their positions. In navigation and geography, it might be used to find a point that is equidistant from two known points, providing a simplified model for locating central facilities or determining mid-travel points.
Yes, the Distance Formula is widely used to find the length of a line segment within various geometric shapes, such as triangles, rectangles, and polygons, in coordinate geometry. Given the coordinates of the endpoints of the line segment, the Distance Formula calculates its length, which can be utilised to explore various properties of the shape, such as perimeter, area, and internal angles. For instance, in a triangle, the Distance Formula can be used to find the lengths of its sides, which can further be used to classify the triangle (scalene, isosceles, or equilateral) and to calculate its area and perimeter.
The Midpoint Formula essentially finds the arithmetic mean (or average) of the x-coordinates and y-coordinates of two points, which gives us a new point that is equidistant from the original two points. The concept of average is used to find a value that represents the centre or middle of a set of values. In the context of the Midpoint Formula, the averages of the x-coordinates and y-coordinates give us a point that is centrally located along the line segment connecting the two original points, thereby ensuring that it is equidistant from them, maintaining the essence of being a midpoint.
Understanding the derivations of the Distance and Midpoint Formulas is crucial as it provides insight into why these formulas are structured the way they are and how they relate to fundamental mathematical concepts, such as the Pythagorean Theorem for the Distance Formula. Grasping the derivations enhances problem-solving skills, as it allows students to logically work through problems and adapt to variations of problems or similar concepts. It also fosters a deeper understanding of the subject, enabling students to apply the concepts in various contexts, including real-world applications and advanced mathematical problems, thereby enhancing their analytical and critical thinking skills.
The Distance Formula can be extended to three-dimensional space to find the distance between two points P(x1, y1, z1) and Q(x2, y2, z2) in a three-dimensional coordinate system. The formula in three-dimensional space is: d = sqrt((x2 - x1)2 + (y2 - y1)2 + (z2 - z1)2). This formula is derived similarly to the two-dimensional Distance Formula but incorporates the z-coordinates to account for the additional dimension. The term (z2 - z1)2 represents the square of the vertical distance between the two points. This formula is particularly useful in fields such as physics and computer graphics where calculations often involve three-dimensional space.
Practice Questions
The distance between two points A(x1, y1) and B(x2, y2) can be found using the Distance Formula: d = sqrt((x2 - x1)2 + (y2 - y1)2). Substituting the given points A(2, 3) and B(6, 7), we get: d = sqrt((6 - 2)2 + (7 - 3)2) d = sqrt(42 + 42) d = sqrt(16 + 16) d = sqrt(32) d = 4sqrt(2).
The midpoint M of a line segment AB can be found using the Midpoint Formula: M = ((x1 + x2)/2, (y1 + y2)/2). Substituting the given points, we get: M = ((2 + 6)/2, (3 + 7)/2) M = (4, 5).
To find the equation of the perpendicular bisector of CD, we first need to find the midpoint M of CD using the Midpoint Formula: M = ((x1 + x2)/2, (y1 + y2)/2). Substituting the given points C(1, 2) and D(7, 8), we get: M = ((1 + 7)/2, (2 + 8)/2) M = (4, 5).
Next, we find the slope mCD of CD using the formula: mCD = (y2 - y1)/(x2 - x1). Substituting the given points, we get: mCD = (8 - 2)/(7 - 1) mCD = 6/6 mCD = 1.
The slope m_PB of the perpendicular bisector is the negative reciprocal of mCD: mPB = -1/mCD mPB = -1.
Using the point-slope form of a line equation y - y1 = m(x - x1) and substituting M(4, 5) and mPB = -1, we get: y - 5 = -1(x - 4) y - 5 = -x + 4 y = -x + 9.
Thus, the equation of the perpendicular bisector of CD is y = -x + 9.
