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Distance in the Coordinate Plane

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Presentation on theme: "Distance in the Coordinate Plane"— Presentation transcript:

1 Distance in the Coordinate Plane
Boardworks High School Geometry (Common Core) Distance in the Coordinate Plane Distance in the Coordinate Plane © Boardworks 2012

2 Information Boardworks High School Geometry (Common Core)
Distance in the Coordinate Plane Information © Boardworks 2012

3 The Cartesian coordinate system
Boardworks High School Geometry (Common Core) Distance in the Coordinate Plane The Cartesian coordinate system The Cartesian coordinate system is named after the French mathematician René Descartes (1596 – 1650). Points in the (x, y) plane are defined by their perpendicular distance from the x- and y-axes relative to the origin, O. The coordinates of a point P are written in the form P(x, y). The x-coordinate gives the horizontal distance from the y-axis to the point. Teacher notes Introduce the system of Cartesian coordinates as providing a way to express the geometry of lines, curves and shapes algebraically. Explain that points are normally denoted by a capital letter followed by the coordinates of the point. Use the embedded Flash movie to demonstrate a variety of points in each of the four quadrants and on the axes. Observe that the x-coordinate of the point becomes negative as the point passes to the left of the y-axis and that the y-coordinate becomes negative as it passes below the x-axis. The y-coordinate gives the vertical distance from the x-axis to the point. © Boardworks 2012

4 Reviewing the Pythagorean theorem
Boardworks High School Geometry (Common Core) Distance in the Coordinate Plane Reviewing the Pythagorean theorem The Pythagorean theorem: In a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the legs. The area of the largest square is c × c or c2. c2 The areas of the smaller squares are a2 and b2. c a2 a Teacher notes Use this slide to review the Pythagorean theorem. It is proved in the presentation Proving the Pythagorean Theorem. The Pythagorean theorem can be written as: b b2 c2 = a2 + b2 © Boardworks 2012

5 The distance formula xb – xa yb – ya AB = √(xa – xb)2 + (ya – yb)2
Boardworks High School Geometry (Common Core) Distance in the Coordinate Plane The distance formula What is the distance between two general points with coordinates A(xa, ya) and B(xb, yb)? horizontal distance between the points: xb – xa vertical distance between the points: yb – ya using the Pythagorean theorem, the square of the distance between the points A(xa, ya) and B(xb, yb) is: (xb – xa)2 + (yb – ya)2 taking the square root gives AB: The distance formula: AB = √(xa – xb)2 + (ya – yb)2 © Boardworks 2012

6 Using the distance formula
Boardworks High School Geometry (Common Core) Distance in the Coordinate Plane Using the distance formula The distance formula is used to find the distance between any two points on a coordinate grid. Find the distance between the points A(7,6) and B(3,–2). write the distance formula: AB = √(xa – xb)2 + (ya – yb)2 substitute values: AB = √(7–3)2 + (6–(–2))2 = Teacher notes The distance between any two points can be found by square rooting the sum of the horizontal distance squared and the vertical distance squared. These distances are shown by the dashed lines. = = √80 = 8.94 units (to the nearest hundredth) © Boardworks 2012

7 Finding the midpoint xa + xb ya + yb
Boardworks High School Geometry (Common Core) Distance in the Coordinate Plane Finding the midpoint The midpoint, M, of two points A(xa, ya) and B(xb, yb) is mean of each coordinate of the two points: xa + xb is the mean of the x-coordinates. 2 Teacher notes Talk through the generalization of the result for any two points (xa, ya) and (xb, yb). As for the generalization for the distance between two points, it doesn’t matter which point is called (xa, ya) and which point is called (xb, yb). ya + yb is the mean of the y-coordinates. 2 © Boardworks 2012

8 Boardworks High School Geometry (Common Core)
Distance in the Coordinate Plane Using midpoint The midpoint of the line segment joining the point (–3, 4) to the point P is (1, –2). Find the coordinates of the point P. Let the coordinates of the points P be (a, b). –3 + a 4 + b midpoint formula: = (1, –2) 2 2 equating the x-coordinates: equating the y-coordinates: –3 + a 4 + b = 1 = –2 2 2 Teacher notes The answer can be checked by verifying that (1, –2) is the midpoint of the line segment joining (–3, 4) and (5, –8). Mathematical practices 1) Make sense of problems and persevere in solving them. Students will need to recognize that they need to apply the midpoint formula and then solve for the point. –3 + a = 2 4 + b = –4 a = 5 b = –8 The coordinates of the point P are (5, –8). © Boardworks 2012


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