/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 6 Calculate the distance between t... [FREE SOLUTION] | 91Ó°ÊÓ

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Calculate the distance between the given points, and find the midpoint of the segment joining them. $$(-2,12) \text { and }(7,15)$$

Short Answer

Expert verified
Distance: \\(3\sqrt{10}\\), Midpoint: \\(\left(\frac{5}{2}, \frac{27}{2}\right)\\).

Step by step solution

01

Calculate the distance using the distance formula

To find the distance between two points \((x_1, y_1)\) and \((x_2, y_2)\), use the distance formula: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]Substitute the given points \((-2, 12)\) and \(7, 15)\) into the formula:\[d = \sqrt{(7 - (-2))^2 + (15 - 12)^2}ewline= \sqrt{(9)^2 + (3)^2}ewline= \sqrt{81 + 9}ewline= \sqrt{90}ewline= 3\sqrt{10}\]Thus, the distance is \(3\sqrt{10}\).
02

Find the midpoint using the midpoint formula

To find the midpoint of the segment joining two points \((x_1, y_1)\) and \((x_2, y_2)\), use the midpoint formula:\[\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)\]Substitute the given points \((-2, 12)\) and \(7, 15)\) into the formula:\[\left(\frac{-2 + 7}{2}, \frac{12 + 15}{2}\right)ewline= \left(\frac{5}{2}, \frac{27}{2}\right)\]So, the midpoint is \(\left(\frac{5}{2}, \frac{27}{2}\right)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Distance Formula
Calculating distances between two points on a flat plane is a fundamental aspect of geometry, and this is made easy using the Distance Formula. The Distance Formula allows us to determine how far apart the points \(x_1, y_1\) and \(x_2, y_2\) are from each other. This formula is essentially derived from the Pythagorean theorem, as it considers the horizontal and vertical distances between the two points as the legs of a right triangle, while the distance between the points is the hypotenuse. To use the formula, apply the following steps:
  • Substitute the coordinates of the two points into the formula: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
  • The expression \(x_2 - x_1\) calculates the horizontal distance between the points.
  • The expression \(y_2 - y_1\) calculates the vertical distance between the points.
  • Square each of these differences and add them together.
  • Finally, take the square root of the sum to find the distance.
Imagine your points are places on a map, and this distance tells you exactly how far one location is from another, no matter which direction you go.
Midpoint Formula
Finding the midpoint of two points on a coordinate plane is like locating the halfway mark between those points. The Midpoint Formula provides a straightforward technique to pinpoint the location that lies exactly in the middle of the segment joining the two points. To calculate the midpoint of a segment connecting \(x_1, y_1\) and \(x_2, y_2\), use the formula:\[ \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) \]Here's how it's done:
  • Add the x-coordinates of both points and divide the sum by 2 to get the x-coordinate of the midpoint.
  • Add the y-coordinates of both points and divide the sum by 2 to get the y-coordinate of the midpoint.
The midpoint is a crucial concept in geometry, reflecting the center of a line segment, just like a median splits a piece of fabric evenly in two. It is widely used in various fields, such as computer graphics, physics, and engineering, helping to determine balance points and central locations.
Coordinate Geometry
Coordinate geometry, sometimes referred to as analytic geometry, merges algebra and geometry using a coordinate system. Think of it like a bridge between geometric shapes and algebraic equations. In this system, every point is described using a pair of numerical coordinates, \(x, y\), which show its position on a plane. This connection between algebra and geometry allows us to use equations to describe geometric figures. Whether it’s finding the distance between points using the Distance Formula, or finding the midpoint of a line segment using the Midpoint Formula, coordinate geometry provides a language to quantify and analyze shapes and positions efficiently.Some exciting aspects include:
  • Determining geometric properties such as the length and midpoint of line segments.
  • Analyzing figures like lines, circles, and parabolas through equations.
  • Solving real-world problems by converting geometric issues into algebraic equations.
Through coordinate geometry, we gain powerful tools to solve geometrical problems with precision and clarity, making it an indispensable part of modern mathematics.

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Most popular questions from this chapter

Two boats leave Key West at noon. The smaller boat is traveling due west. The larger boat is traveling due south. The speed of the larger boat is 10 mph faster than that of the smaller boat. At 3 P.M. the boats are 150 miles apart. Find the average speed of each boat. (Assume there is no current.)

Solve the equation using factoring by grouping: \(x^{3}+2 x^{2}-x-2=0\)

Refer to the following: Einstein's special theory of relativity states that time is relative: Time speeds up or slows down, depending on how fast one object is moving with respect to another. For example, a space probe traveling at a velocity \(v\) near the speed of light \(c\) will have "clocked" a time \(t\) hours, but for a stationary observer on Earth that corresponds to a time \(t_{0} .\) The formula governing this relativity is given by $$ t=t_{0} \sqrt{1-\frac{v^{2}}{c^{2}}} $$ If the time elapsed on a space probe mission is 5 years but the time elapsed on Earth during that mission is 30 years, how fast is the space probe traveling? Give your answer relative to the speed of light.

Solve the equation \(x^{2}-2 x=-2\) by first writing in standard form and then using the quadratic formula. Now plot both sides of the equation in the same viewing screen \(\left(y_{1}=x^{2}-2 x \text { and } y_{2}=-2\right) .\) Do these graphs intersect? Does this agree with the solution set you found?

Explain the mistake that is made. Solve the equation \(x=\sqrt{x+2}\) Solution: $$ x^{2}=x+2 $$ $$ \begin{aligned} x^{2}-x-2 &=0 \\ (x-2)(x+1) &=0 \\ x=-1, x &=2 \end{aligned} $$ This is incorrect. What mistake was made?

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