2.1 The Distance and Midpoint Formulas The distance d(A,B) between two points A(x1,y1) and B(x2,y2) is given by d(A, B) =
( x2 x1 ) 2 ( y2 y1 ) 2
Example: Find the distance between the points A( -2, 3) and B(-5, 0). Here we can choose x1= -2, y1 =3, x2= - 5, y2= 0. Then d(A, B) =
( x2 x1 ) 2 ( y2 y1 ) 2 (5 (2)) 2 (0 3) 2 9 9 18 3 2
Caution: Remember that
a b is not the same as
a b
The midpoint, M, of the segment with endpoints A(x1,y1) and B(x2,y2) has the coordinates x x y y2 M= 1 2, 1 2 2 Example: Find the midpoint of the line segment with endpoints at A( -2, 3) and B(-5, 0). x x y y2 2 (5) 3 0 7 3 M= 1 2 , 1 , , 2 2 2 2 2 2 Caution: The coordinates of M must be written using parentheses.
2.2 Graphs of equations in two variables An equation in two variables is an equation that contains two variables A solution of an equation in two variables is a pair of numbers that satisfies the equation, that is, it makes the equation true. The graph of an equation in two variables is the set of all points (x,y) in the coordinate system such that the pair (x,y) is a solution of the equation. Example : Check whether the points (0,1), ( 2 ,4) are on the graph of 4x2 – y2 = 4 - Is point (0,1) on the graph? We plug in x = 0 and y = 1 into the equation and get 402 -12 = 4, which becomes -1 = 4. This equation is false, therefore the point (0,1) is not a solution. So, (0,1) is not on the graph of this equation. - Is point ( 2 ,2) on the graph? Substituting x with 2 and y with 2 we get 4( 2 )2- 22 = 4, which becomes 8 - 4 = 4. This statement is true, therefore ( 2 ,2) is a solution, and ( 2 ,2) is on the graph of this equation. We often graph an equation by plotting the points, which are the solutions of the equation. To find the points we make the table x |y in which we choose some values of x and find the y so that the pair (x,y) is a solution of the equation. You must plot as many points as necessary to see a pattern (at least 5)
Example: Graph the equation y = x2 + 2 x -3 -2 -1 0 1 2 3
Make the table y= x2+ 2 (-3)2+ 2=11 (-2)2+ 2= 6 (-1)2 + 2= 3 02 + 2 = 2 12 + 2 = 3 22+2 = 6 32+ 2=11
(x,y) (-3,11) (-2,6) (-1,3) (0,2) (1,3) (2,6) (3,11)
-Plot the point and connect them with a continuous curve
The points that are relatively easy to find and plot are x- and y -intercepts. The x-intercepts are the points where the graph crosses (or touches) the x-axis. To find the x-intercepts (i) make y = 0 in the equation (ii) solve the resulting equation for x. If x = a is a solution, then (a,0) is an x-intercept The y-intercepts are the points where the graph crosses (or touches) the y-axis. To find the y-intercepts (i) set x = 0 in the equation (ii) solve the resulting equation for y. If y = b is a solution , then (0,b) is a y-intercept. Example : Find the intercepts of the graph of the equation x- intercepts: (i) set y = 0 in the equation x2 + 02 –x = 6 x2 – x – 6 = 0 (ii) Solve x2 – x – 6 = 0 (x - 3)(x+2) = 0 x-3 = 0 or x + 2 = 0 x=3 or x = -2
x2 + y2 – x =6
x-intercepts: (3,0), (-2, 0) y-intercepts : (i) Set x = 0 in the equation 02 + y2 –0 = 6 y2 = 6 (ii) Solve y2 = 6 y= 6 y-intercepts: (0, 6 ), (0, 6 )
Sometimes the graph is symmetric about the x-axis, or the y-axis or the origin. Advanced knowledge of such property can simplify graphing. Symmetry with respect to x-axis: A graph is symmetric with respect to x-axis, if for every point (x,y) on the graph , the point (x,-y) is also on the graph
To check whether the graph of an equation is symmetric with respect to x-axis (i) replace y by (-y) in the equation and simplify (ii) If the result is the same as the original equation, then the graph is symmetric with respect to the x-axis.
Symmetry with respect to y-axis: A graph is symmetric with respect to y-axis, if for every point (x,y) on the graph , the point (-x,y) is also on the graph
To check whether the graph of an equation is symmetric with respect to y-axis (i) replace x by (-x) in the equation and simplify (ii) If the result is the same as the original equation, then the graph is symmetric with respect to the y-axis.
Symmetry with respect to the origin: A graph is symmetric with respect to the origin, if for every point (x,y) on the graph , the point (-x,y) is also on the graph
To check whether the graph of an equation is symmetric with respect to the origin (i) replace x by (-x) and y by (-y) in the equation and simplify (ii) If the result is the same as the original equation, then the graph is symmetric with respect to the origin. Example: Test the following equation for symmetry 3x a) y 2 x 9 Symmetry with respect to x-axis: (i) Replace y with (-y)
y
3x x 9 3x y 2 x 9 2
(ii)
The equation is not the same as the original. Therefore, the graph is NOT symmetric with respect to x-axis Symmetry with respect to y-axis: (i) Replace x with (-x) 3( x) y x 2 9 3x y 2 x 9 (ii)
The equation is not the same as the original. Therefore, the graph is NOT symmetric with respect to y-axis Symmetry with respect to the origin: (i) Replace x with (-x) and y with (-y) y 3(2x) ( x) 9 3x y 2 x 9 3x y 2 x 9 (ii)
The equation is the same as the original. Therefore, the graph is symmetric with respect to the origin 2.3 Straight Lines
A linear equation in two variables is the equation of the form ax + by = c or y = mx +b. The graph of a linear equation is a straight line. To graph a straight line we need only two points, therefore, to graph a linear equation you must make the table consisting of only two points. If an equation is in the form y = mx +b, you can use x = 0 and x = 1 (or a suitable number that would make the calculations easy); for an equation in the form ax + by = c, you can choose x- and y- intercepts, that is points for which y and x, respectively, is 0. Remark: Use a straight edge to graph the lines. Points on the tests will be taken off for sloppy and inaccurate graphs!
Example: Graph y 3 x 2 5
x 0 5
y 2 3+2 = 5
Example: Graph 2x + 6y = 18 x y 0 3 9 0
Example: Graph y = 4 x y 0 4 1 4
Example: Graph x = -3 x y 0 dne 1 Dne -3 0 -3 1
The slope of the line passing through the points A(x1,y1) and B(x2,y2) , x1 ≠ x2 is defined as m
y 2 y1 change in y y rise x2 x1 change in x x run
The slope of the line is the same regardless of the choice of points A, B Remarks: a) If a line is horizontal, its slope is 0 (since y2-y1 = 0) b) If a line is vertical, its slope is not defined (since x2-x1=0) c) The quantity y is called the average rate of change of y with respect to x. x
Example: Find the slope of the line passing through A(-5,3) and B(-1, -4) Use the formula m
y 2 y1 43 7 x2 x1 1 (5) 4
The slope of the line measures the steepness of the line. The larger the absolute value of the slope the steeper the line.
If the slope m is positive, the line is rising from left to right. If the slope is negative , than the line is falling from left to right. If the slope is zero, then the line is horizontal If the slope is undefined, then the line is vertical.
Example: Find the slope of the line given by the equation y = mx +b. To find a slope we need two points: x y 0 b 1 m+b With these points the slope is y 2 y1 m b b m m x2 x1 1 0 1
Therefore, in the equation y = mx +b, m is the slope and b is the y-intercept. Equation of the line in the form y = mx +b is said to be in the slope-intercept form. Example: Find the slope and the y-intercept of the line given by 2x -3y = 9 It is enough to rewrite this equation in the from y = mx +b, that is, to solve for y 2x-3y = 9 -3y = -2x + 9 y
2x 9 2 x 3 3 3 3
Therefore, the slope m= 2/3 and the y-intercept is -3 Graphing a line using the slope m and a point A(x1, y1) One point, A, on the line is already given, so to graph this line we need to find one more point. We can do this using the definition of the slope m = rise/run and the fact that the slope is constant for a line and does not depend on the choice of points. Thus, if m = p/q, where q is positive, then - Starting from the given point A move vertically p units up if p is positive or p units down, if p is negative - Move horizontally to the right by q units - The ending point, B, belongs to the line - Use the straight edge to graph the line through A and B Example: Graph the line with the slope m=2/3 and y-intercept -3 A point here is (0,-3) and the slope m = 2/3. We choose rise = 2 and run = 3
Equation of the line given the slope m and a point A(x1,y1):
y-y1 = m(x-x1)
Example: Find the equation, in the slope intercept form, of the line with slope m= -2/3 passing through the point A( -1, 3) (i) Determine m, x1, y1 m = -2/3, x1 = -1 and y1 = 3 (ii) Use the formula y y1 m( x x1 ) 2 y 3 ( x (1)) 3
(iii)
Remove parentheses and write in y = mx +b form 2 y 3 ( x 1) 3 2 2 y x 3 3 3 2 7 y x 3 3
Equation of the line passing through two points A(x1,y1) and B(x2,y2) To find an equation of the line : (i)
Find the slope m m
(ii)
y 2 y1 x 2 x1
Use point A (or B) and the slope to write the equation of the line
Example: Find the slope of the line passing through A(-1,3) and B(-4,5) y 2 y1 53 2 2 x2 x1 4 (1) 3 3
(i)
m
(ii)
Use A(-1,3) and m to write the equation in the slope intercept form 2 y 3 ( x 1) 3 2 2 y x 3 3 3 2 7 y x 3 3
Example: Find an equation, in slope intercept form if possible, of the line with undefined slope and passing through A(-1,3) (i) Slope m is undefined, so the line is vertical (ii) Its equation is x= x-coordinate of the point, that is x = -1 In general, the equation of the vertical line passing through A(x1, y1) is x = x1 Example: Find the equation of a horizontal line passing through A(-1,3) (i) Line is horizontal, so m= 0 (ii) Equation is y – 3 =0(x+1) or y =3 In general the equation of the horizontal line passing through A(x1, y1) is y = y1 Parallel lines Two nonvertical lines are parallel if and only if their slopes are the same and their y-intercepts are different. Example: Find the equation, in slope-intercept form if possible, of the line parallel to the line 2x -3y = 9 and passing through A(-1,3) (i) Find the slope of the given line 2x-3y = 9
-3y = -2x + 9 2x 9 2 m = 2/3 y x 3 3 3 3 (ii)determine the slope of the parallel line: m || = m = 2/3 (iii)
Use the point A and the slope m|| to write the equation 2 ( x 1) 3 2 2 y x 3 3 3 2 11 y x 3 3 y 3
Perpendicular lines Two (nonvertical, nonhorizontal) lines are perpendicular if and only if the product of their slopes is equal to -1, or, if one slope is the negative reciprocal of the other. Example: Find the equation, in slope-intercept form if possible, of the line perpendicular to the line 2x -3y = 9 and passing through A(-1,3) (i) Find the slope of the given line 2x-3y = 9 -3y = -2x + 9 2x 9 2 m = 2/3 y x 3 3 3 3 1 3 (ii)determine the slope of the perpendicular line: m = m 2 (i) Use the point A and the slope m to write the equation
3 y 3 ( x 1) 2 3 3 y x 3 2 2 3 3 y x 2 2 Remarks: a) Two vertical lines are parallel. Two horizontal lines are parallel. b) A line perpendicular to a vertical line is horizontal; a line perpendicular to a horizontal is vertical.
2.4 Circles A circle with center at C and the radius r is the set of all points (x, y) whose distance from C is r.
The equation of a circle with center at C(h,k) and radius r is therefore: (x-h)2 + (y-k)2 = r2 This equation is called the standard equation of the circle. If C is the origin (0,0), then the circle with radius r has the equation x 2 + y2 = r2 The circle with center at origin and radius r =1 (x2 + y2 = 1) is called the unit circle. Example: Write the equation of the circle with center C(-2,5) and radius 3 (i) Identify h, k, r h = -2, k = 5, r = 3 (ii) Use the standard form (x-h)2 + (y-k)2 = r2 to write the equation (x-(-2))2 +(y-5)2 = 32 (x+2)2 + (y-5)2 = 9 To graph a circle with center C(-2,5) and radius r = 3 (i) Plot the center C(-2, 5) (ii) Find 4 points that are r = 3 units away from the center and are above, below, to the left and to the right of the center. Use them to graph a circle.
To find the intercepts of a circle, follow general procedure of finding the intercepts of the graph of the equation in two variables (sec 2.2)
Sometimes an equation of a circle can be written as x2 + y2 + ax +by +c = 0 (when a2 + b2 -4c >0), if we remove parentheses in the standard equation. To find the center and the radius of a circle given by x2 + y2 + ax +by +c = 0 we rewrite it in the standard form. The procedure is outlined in the following example Example: Find the center and the radius of the circle given by x 2 + y2 + 4x -6 y +2 = 0 (i) Group x’s and y’s together and move the constant 2 to the right hand side (x2 + 4x)+ (y2 – 6y)= -2 (ii) Complete each parentheses to a perfect square, that is add a suitable number to each expression in parentheses, so it can be factored and written as (x-h)2 and (y-k)2, respectively (x2 + 4x + (4/2)2) + (y2 – 6x + (6/2)2) = -2 + (4/2)2 + (6/2)2 (x2 + 4x +4) + (y2 -6y +9) = -2 + 4 + 9 (x+2)2 + (y -3)2 = 11 (x-(-2))2 + (y – 3)2 = 11 (iii) Read the h, k, r from the equation: h = -2, k = 3, r2 = 11, so r = 11
