Analytic Geometry 1 -1 Cartesian Coordinate System - PDF document

Analytic Geometry ةيليلحتلا ةسدنھلا 1 -1 Cartesian Coordinate System يتراكيدلا تايثادحلئا ماظن • The Cartesian coordinate system, or the rectangular coordinate system, is a geometrical system that is used to determine the locations of points in a plane . • Points are located with respect to a reference point called the origin which is the intersection point of a horizontal line, known as x -axis, and a vertical line called y -axis . • The x and y axes divide the Cartesian plane into four regions called quadrants . • Each point in the plane is defined by an ordered pair ( x , y ) of real numbers called the coordinates of the point . • An example of ordered pairs or coordinates is the point P below: 2 GFP - Sohar University SET 2 - Chapter 1

SET 2 - Chapter 1 3 GFP - Sohar University 1 - 2 The Distance Formula ةفاسملا داجيإ نوناق • The distance d between two points A ( x 1 , y 1 ) and B ( x 2 , y 2 ) can be found from the distance formula: 4 GFP - Sohar University SET 2 - Chapter 1

Example 1: Find the distance between the points C (3, ‒ 4) and D ( ‒ 13, ‒ 11). Solution: SET 2 - Chapter 1 5 GFP - Sohar University 1 - 3 The Midpoint Formula فصتنملا ةطقن داجيإ نوناق • The coordinates of the midpoint of a line segment joining the two points A ( x 1 , y 1 ) and B ( x 2 , y 2 ) are found by averaging the coordinates of the endpoints. • The midpoint formula is : 6 GFP - Sohar University SET 2 - Chapter 1

Example 2: F is the midpoint between points C (3, ‒ 4) and D ( ‒ 13, ‒ 11). Find its coordinates. Solution: SET 2 - Chapter 1 7 GFP - Sohar University 1 - 4 The Slope of a Line ميقتسملا طخلا ليم • The slope of a line is a measurement of its steepness and direction. • Slope of a line m is calculated from the following formula which is called the slope formula: 8 GFP - Sohar University SET 2 - Chapter 1

• Depending on the direction of the line, its slope could be positive, negative, zero or undefined and as shown below. SET 2 - Chapter 1 9 GFP - Sohar University Example 3: Find the slope of the line that passes through points P ( ‒ 4, 8) and R (9, ‒ 7). Solution: 10 GFP - Sohar University SET 2 - Chapter 1

Example 4: Find the slope of the lines a , b , c and d shown in the figure below. Solution: SET 2 - Chapter 1 11 GFP - Sohar University 1 - 5 Parallel Lines ةيزاوتملا طوطخلا 12 GFP - Sohar University SET 2 - Chapter 1

1 - 6 Perpendicular Lines ةدماعتملا طوطخلا SET 2 - Chapter 1 13 GFP - Sohar University Example 5: Lines m and n are parallel. If the slope of line m is ‒ 0.48, what is the slope of line n ? Solution: Since the two lines are parallel, then they have the same slope. So, the slope of line n = ‒ 0.48 Example 6: Line c is perpendicular to line d and the slope of line c is 0.5. Find the slope of line d . Solution: Since lines c and d are perpendicular, then their slopes are opposite reciprocals of one another. Therefore, the slope of line d = 14 GFP - Sohar University SET 2 - Chapter 1

1 - 7 x -Intercept and y -Intercept يداصلا عطقملا و ينيسلا عطقملا • For a non-horizontal line, x -intercept is the x -coordinate of the point where the line intersects x -axis. • In the same way, for a non-vertical line, y -intercept is the y - coordinate of the point where it intersects y -axis. a SET 2 - Chapter 1 15 GFP - Sohar University 1 - 8 Equations of Lines ميقتسملا طخلا ةلداعم • The equation of a line is a mathematical sentence that describes the relationship between the x -coordinate and the y -coordinate of all its points . • The equation of a line is of the first degree and is therefore called a linear equation . • Straight line equation may be written in any of the following three forms : where m is the slope. where m is the slope, and b is the y -intercept. 16 GFP - Sohar University SET 2 - Chapter 1

Example 7: Draw the graph of the line whose equation is 2 x ‒ 3 y = 6 using two randomly selected points. Solution: Let x = 1 , then: Let y = 1 , then: Thus, (1, ‒ 1.33) is the first point. Thus, (4.5, 1) is the second point. SET 2 - Chapter 1 17 GFP - Sohar University The graph of 2 x ‒ 3 y = 6 is as shown in the figure below. 18 GFP - Sohar University SET 2 - Chapter 1

Example 8: Draw the graph of the line whose equation is 2 x ‒ 3 y = 6 using the x -intercept and the y -intercept. Solution: x -intercept  y = 0 So, (3, 0) is the first point. y -intercept  x = 0 Therefore, (0, ‒ 2) is the first point. SET 2 - Chapter 1 19 GFP - Sohar University Example 9: Draw the graph of : (a) x = 4 (b) y = ‒ 2 Solution: (a) The graph of x = 4 is a vertical line (b) The graph of y = ‒ 2 is a horizontal line with x -coordinate = 4 for all its points with y -coordinate = ‒ 2 for all its points 20 GFP - Sohar University SET 2 - Chapter 1

Example 10: Find the equation of the line that passes though the points (4, –5) and (–11, 3). Write the equation in point-slope form, standard form and slope-intercept form. Solution: Find the slope first: Write the equation in point-slope form: SET 2 - Chapter 1 21 GFP - Sohar University To write the equation in the standard form, multiply both sides of the equation by 15:  Rearrange the equation 8 x + 15 y = ‒ 43 to write it in the slope-intercept form: Divide both sides by 15:  22 GFP - Sohar University SET 2 - Chapter 1

Example 11: Determine whether the lines 6 x + 4 y = ‒ 9 and 8 x ‒ 12 y = ‒ 7 are parallel or perpendicular or neither. Solution: Find the slopes of the two lines and compare them:  m 1 =  m 2 = So, the two lines are perpendicular since their slopes are opposite reciprocals of one another. SET 2 - Chapter 1 23 GFP - Sohar University Example 12: Which of the points A (2, 1.6) and B (1, –2.2) lie on the graph of the lin e 3 x + 5 y = 14? Solution: If a point lies on the graph of a line then it satisfies its equation. Check point A (2, 1.6):  Point A (2, 1.6) lies on the graph of 3 x + 5 y = 14 Check point B (1, –2.2) :  Point B (1, –2.2) doesn’t lie on the graph of 3 x + 5 y = 14 24 GFP - Sohar University SET 2 - Chapter 1

1 - 9 Equations of Circles ةرئادلا ةلداعم • The radius r of a circle with a centre at the point ( h , k ) can be found using the distance formula between the centre and any point on the circle ( x , y ) and as follows: • The standard form of the equation of a circle of radius r with centre at the point ( h , k ) is: ( x – h ) 2 + ( y – k ) 2 = r 2 SET 2 - Chapter 1 25 GFP - Sohar University Example 13: The point (3, 4) lies on a circle whose centre is at ( ‒ 1, 2), as shown in the Figure below. Write the standard form of the equation of this circle. Solution: The radius r of the circle is the distance between ( ‒ 1, 2) and (3, 4): Using ( h , k ) = ( ‒ 1, 2) and r , The equation of the circle is: Standard Form 26 GFP - Sohar University SET 2 - Chapter 1

1 - 10 Symmetry of Equations تلبداعملا رظانت • A graph has symmetry with respect to the y -axis if whenever ( x , y ) is on the graph, so is the point ( ‒ x , y ). • A graph has symmetry with respect to the origin if whenever ( x , y ) is on the graph, so is the point ( ‒ x , ‒ y ). • A graph has symmetry with respect to the x -axis if whenever ( x , y ) is on the graph, so is the point ( x , ‒ y ). SET 2 - Chapter 1 27 GFP - Sohar University Example 14: Test y = x 2 + 2 for symmetry with respect to the x -axis, the y -axis, and the origin . . Solution: x -Axis y -Axis We replace y with ‒ y : We replace x with ‒ x : Multiplying both sides by ‒ 1: Simplifying gives: The resulting equation is not The resulting equation is equivalent to equivalent to the original equation, the original equation, so the graph is so the graph is not symmetric with symmetric with respect to the y -axis. respect to the x -axis. 28 GFP - Sohar University SET 2 - Chapter 1

Origin We replace x with ‒ x and y with ‒ y : Simplifying gives: The resulting equation is not equivalent to the original equation, so the graph is not symmetric with respect to the origin. SET 2 - Chapter 1 29 GFP - Sohar University Example 15: Test x 2 + y 4 = 5 for symmetry with respect to the x -axis, the y -axis, and the origin . . Solution: x -Axis y -Axis We replace y with ‒ y : We replace x with ‒ x : The resulting equation is equivalent The resulting equation is equivalent to to the original equation, so the the original equation, so the graph is graph is symmetric with respect to symmetric with respect to the y -axis. the x -axis. 30 GFP - Sohar University SET 2 - Chapter 1

Origin We replace x with ‒ x and y with ‒ y : The resulting equation is equivalent to the original equation, so the graph is symmetric with respect to the origin. SET 2 - Chapter 1 31 GFP - Sohar University