Problem 16
Question
Find the distance between the two points. Round the result to the nearest hundredth if necessary. $$(3,-1),(0,3)$$
Step-by-Step Solution
Verified Answer
The distance between the two points (3,-1) and (0,3) is 5.
1Step 1: Substitute Values
Substitute the given values for \(x_1\), \(y_1\), \(x_2\), and \(y_2\) into the distance formula. That gives us: \(d = \sqrt{(0 - 3)^2 + (3 - (-1))^2}\).
2Step 2: Simplify Inside The Square Root
Simplify inside the square root. This becomes: \(d = \sqrt{(-3)^2 + (4)^2} = \sqrt{9 + 16}\).
3Step 3: Compute The Distance
Continue to carry out the calculations: \(d = \sqrt{25} = 5\).
Key Concepts
Coordinate GeometryPythagorean TheoremSquare RootSimplifying Expressions
Coordinate Geometry
Understanding coordinate geometry is essential when analyzing positions and shapes on a two-dimensional plane. To visualize it, imagine a grid with horizontal and vertical lines crossing at a central point called the origin. This grid is divided into four sections, known as quadrants, allowing us to pinpoint any location with a pair of numbers: the x-coordinate and the y-coordinate.
When we refer to coordinates, such as \( (3,-1) \) and \( (0,3) \), each pair represents a specific point on this grid: the first number is the x-coordinate (distance from the vertical axis), and the second is the y-coordinate (distance from the horizontal axis). By knowing these positions, we can use other mathematical tools, such as the distance formula, to calculate distances directly on this coordinate plane.
When we refer to coordinates, such as \( (3,-1) \) and \( (0,3) \), each pair represents a specific point on this grid: the first number is the x-coordinate (distance from the vertical axis), and the second is the y-coordinate (distance from the horizontal axis). By knowing these positions, we can use other mathematical tools, such as the distance formula, to calculate distances directly on this coordinate plane.
Pythagorean Theorem
At the heart of coordinate geometry, especially when determining distance, lies the Pythagorean Theorem. This theorem applies to right-angled triangles, asserting that the square of the hypotenuse (longest side) is equal to the sum of the squares of the other two sides. In formula terms, if 'a' and 'b' are the legs of a triangle and 'c' is the hypotenuse, then \( a^2 + b^2 = c^2 \).
How does this relate to our distance problem? When you plot points on a graph and want to find the distance between them, you essentially form a right-angled triangle with these points and their projections onto the x- and y-axes. Hence, to find the distance (which would be the hypotenuse in our scenario), we utilize this powerful theorem.
How does this relate to our distance problem? When you plot points on a graph and want to find the distance between them, you essentially form a right-angled triangle with these points and their projections onto the x- and y-axes. Hence, to find the distance (which would be the hypotenuse in our scenario), we utilize this powerful theorem.
Square Root
After applying the Pythagorean Theorem, we often encounter the square root symbol in the final step of our calculation. It represents a fundamental arithmetic operation. The square root symbol \( \sqrt{\cdot} \) asks the question, 'What number times itself gives me the number inside the symbol?' So, if you see \( \sqrt{25} \) in a math problem, you're essentially looking for a number that, when multiplied by itself, gives 25. The answer, in this case, is 5.
Understanding square roots is crucial when finalizing the distance between two points, as it's the last operation performed in the formula that provides us with the actual measure of distance we're seeking.
Understanding square roots is crucial when finalizing the distance between two points, as it's the last operation performed in the formula that provides us with the actual measure of distance we're seeking.
Simplifying Expressions
Simplifying expressions is a fundamental skill in mathematics, allowing us to transform complex equations into simpler forms that are easier to interpret and solve. When working with the distance formula, simplifying involves squaring numbers, adding or subtracting them, and then finding the square root of the result.
Example:
For the points \( (3,-1) \) and \( (0,3) \), we start with the raw expression \( \sqrt{(-3)^2 + (4)^2} \) and then simplify inside the square root to get \( \sqrt{9 + 16} \) which then further simplifies to \( \sqrt{25} \) and finally, to the distance value of 5. The simplification makes it clear that \( (-3)^2 \) and \( (4)^2 \) are simply the square of the x-difference and the y-difference between our points, respectively. This step-by-step simplification is critical for students to follow and understand how each part of the expression contributes to the final distance.Other exercises in this chapter
Problem 16
Solve the equation. Check for extraneous solutions. $$\sqrt{x}-1=0$$
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Prove the theorem. (Use the basic axioms of algebra and the definition of subtraction given in Example \(1 .\) ) If \(a, b,\) and \(c\) are real numbers, then \
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Find the missing length of the right triangle if a and b are the lengths of the legs and c is the length of the hypotenuse. $$a=10, b=24$$
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Sketch the graph of the function. $$y=\sqrt{x}+5$$
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