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Interactive Audio Lesson
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Understanding the Distance Between Points on Axes
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Today we will start by understanding how to measure distances between points on the Cartesian plane. For instance, if we have points A and B on the x-axis, how would we deduce the distance?
We would subtract the x-coordinates, right? Like if A is (4,0) and B is (6,0), the distance is |6 - 4| = 2 units.
Exactly! Now, what about when we have points on the y-axis?
We would do the same. For points C(0,3) and D(0,8), it would be |8 - 3| = 5 units.
Great understanding! This lays the groundwork for understanding distance in two dimensions. Let's dive deeper into when the points are not aligned.
Applying Pythagorean Theorem to Calculate Distance
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When we have points in different quadrants or positioned randomly, we will derive our Distance Formula using the Pythagorean theorem. Let's take points P(4, 6) and Q(6, 8); how would you approach finding the distance now?
I think we would need to draw a right triangle to see the lengths of its sides?
Correct! The horizontal distance would be the difference in x-coordinates, and the vertical distance from y-coordinates. It forms the equation PQΒ² = PTΒ² + QTΒ².
So that means PQ = β((6-4)Β² + (8-6)Β²) = β(2Β² + 2Β²) = β8?
Absolutely! Thus, the formula PQ = β((x2 - x1)Β² + (y2 - y1)Β²) allows us to compute distances regardless of the points' positions.
Distance Between Points and Their Practical Applications
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Now that we've covered the Distance Formula, how might it help in identifying if three points form a triangle?
We could calculate the distances between each pair of points and check if the sum of any two distances is greater than the third?
Like using PQR with points P(3,2), Q(-2,-3), R(2,3) to check?
Exactly! After calculating PQ, QR, and PR, you will confirm if it satisfies the triangle inequality theorem.
And that will also tell us if it's a right triangle if any two distances equal the third squared!
Perfect summary! The formula is indeed crucial in geometry and applicable in multiple fields!
Introduction & Overview
Read a summary of the section's main ideas. Choose from Basic, Medium, or Detailed.
Quick Overview
Standard
This section introduces the Distance Formula, derived from the Pythagorean theorem. It explains how to find the distance between two points on the Cartesian plane, regardless of their location in relation to the axes. The formula is essential for various applications in geometry and real life.
Detailed
Distance Formula
In this section, we explore the Distance Formula, an essential tool in coordinate geometry used to calculate the distance between two points in a Cartesian coordinate system. The process involves applying the Pythagorean Theorem to the coordinates of the points in question.
Key Concepts Covered:
- Understanding Points on Axes: The section begins with visualizing points on the x-axis and y-axis, explaining how distances can simply be calculated by subtraction when points are aligned along the same axis.
- Using Pythagorean Theorem: When points are not aligned, the Distance Formula is derived by forming a right triangle from the points and applying the Pythagorean theorem. The formula can be expressed as:
\[ PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
This formula allows us to find the distance between any two points, (x1, y1) and (x2, y2), in a two-dimensional space.
- Examples and Applications: The chapter demonstrates several examples, including scenarios like determining whether points form a triangle or calculating distances between specific points on a grid.
- Special Cases: Special cases are discussed, such as finding the distance from the origin and the unconditional distances between various geometric figures.
By the end of the section, students should be able to apply the Distance Formula effectively, whether for academic exercises or practical applications within fields such as engineering and physics.
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Audio Book
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Understanding the Distance Between Two Points
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Let us consider the following situation:
A town B is located 36 km east and 15 km north of the town A. How would you find the distance from town A to town B without actually measuring it? Let us see. This situation can be represented graphically as shown in Fig. 7.1. You may use the Pythagoras Theorem to calculate this distance.
Detailed Explanation
In this first part, we are introduced to a practical scenario where we need to find the distance between two points (town A and town B) without physically measuring it. The coordinates of point B relative to point A can be described by using the Pythagorean theorem. The theorem states that in a right 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.
Examples & Analogies
Imagine you have a straight road between two towns: one town is further east and also higher up on the map. Instead of walking that road to measure the distance, you can set up a right triangle: one leg represents how far east you need to travel and the other represents how far north. The hypotenuse gives you the direct distance between the two points.
Finding Distance on the Coordinate Axes
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Now, suppose two points lie on the x-axis. Can we find the distance between them? For instance, consider two points A(4, 0) and B(6, 0). The points A and B lie on the x-axis. From the figure you can see that OA = 4 units and OB = 6 units. Therefore, the distance of B from A, i.e., AB = OB β OA = 6 β 4 = 2 units. So, if two points lie on the x-axis, we can easily find the distance between them.
Detailed Explanation
When points lie on the same axis (the x-axis in this case), we can simply take the difference of their x-coordinates to find the distance between them. The coordinates for points A and B are provided as (4, 0) and (6, 0). By subtracting the x-coordinate of A from that of B, we find that the distance AB is 2 units.
Examples & Analogies
Think of a ruler laid horizontally: if you have one tick mark at 4 cm (point A) and another at 6 cm (point B), you can simply count the number of ticks between them to find they are 2 cm apart.
Distance Between Points on the Y-Axis
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Now, suppose we take two points lying on the y-axis. If the points C(0, 3) and D(0, 8) lie on the y-axis, similarly we find that CD = 8 β 3 = 5 units.
Detailed Explanation
The same principle applies when points lie on the y-axis. By subtracting the y-coordinates of points C and D from each other, we can calculate the vertical distance between them. Hence, if C is located at (0, 3) and D at (0, 8), the distance CD is 5 units.
Examples & Analogies
Imagine a vertical line on a graph paper where point C is 3 cm up from the origin and point D is 8 cm up. You measure the space between these two points straight up, which turns out to be 5 cm.
Distance Between Points in Cartesian Plane
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Now, if we consider two points not lying on coordinate axis, can we find the distance between them? Yes! We shall use Pythagoras theorem to do so. Let us see an example. In Fig. 7.3, the points P(4, 6) and Q(6, 8) lie in the first quadrant. How do we use Pythagoras theorem to find the distance between them?
Detailed Explanation
In this example, we need to find the distance between two points that donβt sit on the axes. By using the Pythagorean theorem, we can determine the length of the hypotenuse of a right triangle formed by the points. To visualize, we can drop perpendiculars from P and Q to the x-axis, forming a triangle where we can label the sides accordingly.
Examples & Analogies
Consider navigating in a park where two landmarks are positioned diagonally. To find the shortest path between them, envision drawing a straight line from one landmark to the other and using the right triangle formed with the edges of the paths, just like in the Pythagorean theorem.
Using the Distance Formula
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Now, let us find the distance between any two points P(x1, y1) and Q(x2, y2). Draw PR and QS perpendicular to the x-axis. A perpendicular from the point P on QS is drawn to meet it at the point T. Then, OR = x1, OS = x2. So, RS = x2 β x1 = PT. Also, SQ = y2, ST = PR = y1. So, QT = y2 β y1. Now, applying the Pythagoras theorem in β³PTQ, we get PQΒ² = PTΒ² + QTΒ².
Detailed Explanation
This segment guides us on how to generalize the distance between any two points in the Cartesian plane. By establishing right triangles and applying the Pythagorean theorem, we arrive at the square of the distance formula. The final expressionββ((x2 - x1)Β² + (y2 - y1)Β²)βis the distance formula, which can be applied universally to calculate distances regardless of the position of points.
Examples & Analogies
Think about needing to find the distance between two points on a city map. You can directly measure along the streets (like using coordinate axes) or find the most direct path (hypotenuse) using a measurement tool involved in applying the distance formula.
Special Cases of the Distance Formula
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In particular, the distance of a point P(x, y) from the origin O(0, 0) is given by OP = β(xΒ² + yΒ²).
Detailed Explanation
This chunk highlights a special case of the distance formula, specifically when one of the points is the origin (0, 0). By applying our derived distance formula, it simplifies to the distance from the origin being the square root of the sum of the squares of the coordinates.
Examples & Analogies
Picture a drone taking off from a central point (the origin) and flying to a specific point in a park. To figure out how far it has traveled, we can directly use the coordinates (x, y) of that point and apply the formula.
Definitions & Key Concepts
Learn essential terms and foundational ideas that form the basis of the topic.
Key Concepts
-
Understanding Points on Axes: The section begins with visualizing points on the x-axis and y-axis, explaining how distances can simply be calculated by subtraction when points are aligned along the same axis.
-
Using Pythagorean Theorem: When points are not aligned, the Distance Formula is derived by forming a right triangle from the points and applying the Pythagorean theorem. The formula can be expressed as:
-
\[ PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
-
This formula allows us to find the distance between any two points,
(x1, y1)and(x2, y2), in a two-dimensional space. -
Examples and Applications: The chapter demonstrates several examples, including scenarios like determining whether points form a triangle or calculating distances between specific points on a grid.
-
Special Cases: Special cases are discussed, such as finding the distance from the origin and the unconditional distances between various geometric figures.
-
By the end of the section, students should be able to apply the Distance Formula effectively, whether for academic exercises or practical applications within fields such as engineering and physics.
Examples & Real-Life Applications
See how the concepts apply in real-world scenarios to understand their practical implications.
Examples
-
Example 1: Finding the distance between points A(4, 0) and B(6, 0) results in a distance of 2 units.
-
Example 2: Verifying if points P(3, 2), Q(-2, -3), and R(2, 3) form a triangle using the Distance Formula.
Memory Aids
Use mnemonics, acronyms, or visual cues to help remember key information more easily.
π΅ Rhymes Time
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In the plane, find the spaces, between the points, use their places.
π Fascinating Stories
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Once there were two friends named Point A and Point B, who wanted to measure the distance between them without a ruler. They laid a right triangle down, showing how far apart they were, thanks to the Pythagorean theorem, they calculated their distance and became even closer friends.
π§ Other Memory Gems
-
Base and Height for PQ; Distance is always the positive view.
π― Super Acronyms
D for Distance, P for Points, C for Calculate
- DPC.
Flash Cards
Review key concepts with flashcards.
Glossary of Terms
Review the Definitions for terms.
-
Term: Distance Formula
Definition:
A formula used to determine the distance between two points (x1, y1) and (x2, y2) in the Cartesian plane: \( PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \).
-
Term: Pythagorean Theorem
Definition:
A mathematical principle that relates the lengths of the sides of a right triangle, defined as aΒ² + bΒ² = cΒ².
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Term: Collinear
Definition:
Points that lie on the same straight line.
-
Term: Quadrant
Definition:
One of the four regions created by the intersection of the x-axis and y-axis in a coordinate plane.
Example 4 - Variation
Find a relation between $x$ and $y$ such that the point $(x, y)$ is equidistant from the points $(2, 4)$ and $(6, 8)$.
Solution: Let $(x, y)$ be equidistant from the points $A(2, 4)$ and $B(6, 8)$.
We can set up the equation based on the distance formula:
\[ AP = BP \]
This gives us:
\[ \sqrt{(x - 2)^2 + (y - 4)^2} = \sqrt{(x - 6)^2 + (y - 8)^2} \]
Squaring both sides leads to:
\[ (x - 2)^2 + (y - 4)^2 = (x - 6)^2 + (y - 8)^2 \]
Expanding both sides results in:
\[ (x^2 - 4x + 4 + y^2 - 8y + 16) = (x^2 - 12x + 36 + y^2 - 16y + 64) \]
Thus, simplifying gives:
\[ -4x + 20 - 8y = -12x + 100 - 16y \]
This leads to the simplified equation:
\[ 8x - 8y = 80 \]
Hence, we find:
\[ x - y = 10 \]
Thus, the relation between $x$ and $y$ is $x - y = 10$.
Similar Question
Find a relation between $x$ and $y$ such that the point $(x, y)$ is equidistant from the points $(3, 1)$ and $(7, 5)$.
Solution: Let $(x, y)$ be equidistant from the points $A(3, 1)$ and $B(7, 5)$.
We have:
\[ AP = BP \]
Using the distance formula:
\[ \sqrt{(x - 3)^2 + (y - 1)^2} = \sqrt{(x - 7)^2 + (y - 5)^2} \]
Squaring both sides,
\[ (x - 3)^2 + (y - 1)^2 = (x - 7)^2 + (y - 5)^2 \]
Expanding and simplifying leads us to the relation.