Section 1.1 Linear Equations: Slope and Equations of Lines


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1 Section. Linear Equations: Slope and Equations of Lines Slope The measure of the steepness of a line is called the slope of the line. It is the amount of change in y, the rise, divided by the amount of change in x, the run. Let ( x, y ) and (, ) x y be two arbitrary points on the coordinate plane. The slope of the line that passes through these two points, denoted by m, is given by y y m x x = = = rise run vertical change, horizontal change provided that x x 0. If the slope is positive, the line rises to the right. If the slope is negative, the line falls to the right. If the slope is zero, the line is a horizontal line. If the slope is undefined, the line is a vertical line. (This occurs when x x = 0.) Example : Find the slope of each line shown below. A. B. C. Math Page of 9 Section.
2 Solution: A. To find the slope of the given line, choose two points on the graph with integer coordinates, like ( 0, ) and (, 0), and keep in mind that the slope is rise over run. Let us start at ( 0, ). To get to (, 0 ), we rise units, then run two units. Hence, we get m = =. B. To find the slope of the given line, choose two points on the graph with integer coordinates, like ( 0, ) and, 0, and keep in mind that the slope is rise over run. Let us start at ( 0, ). To get to (, 0 ), we move down four units, then to the right two units. Hence, we get m = =. C. To find the slope of the given line, choose two points on the graph with integer coordinates, like (, ) and (, ), and keep in mind that the slope is rise over run. Let us start at (, ). To get to (, ), we move up unit, then to the left units. Hence, we get m = =. Example : Find the slope of the line that passes through the following points. A. (, ) and (, ) B. (, 0) and ( 6, 7) C. E., 8 and, 6, 6 and, 6 D. (, 9) and, 9 Math Page of 9 Section.
3 Solution: A. Let ( x, y ) represent (, ), and ( x, y ) represent (, ) y y 6 6 m = = = = x x ( ) 7 7. Notice that if we had switched the selection of ( x, y ) and (, ) same result: Let ( x, y ) represent (, ), and ( x, y ) represent (,) 6 6 = = = = 7 7 y y m x x B. Let ( x, y ) represent (, 0), and ( x, y ) represent ( 6, 7) y y m x x 7 0 = = = = 6. x y, we would obtain the. Then C. Let (, ) x y represent, 8, and, x y represent, 6. + y y m = = = x x To simplify the complex fraction, we first simplify the numerator and denominator separately (by finding a common denominator for each) m = = To divide the two fractions, we multiply by the reciprocal of the denominator and then simplify m = = = = Note: Another way to simplify the complex fraction is to multiply both the numerator and denominator by the least common multiple of, 6, and (which is ). Math Page of 9 Section.
4 m = = = D. Let ( x, y ) represent (, 9), and (, ) x y represent, 9. y y m x x = = = 9 9 ( ) 0 We do not need to compute the denominator, since zero divided by a nonzero number is zero. Recall that if the slope of a line is zero, the line is a horizontal line. E. Let (, ) x y represent, 6, and, x y represent, 6. y y m x x Intercepts = = = ( ), which is undefined Division by zero is undefined. Recall that if the slope of a line is undefined, the line is a vertical line. The yintercept is the ycoordinate of the point where the graph crosses the yaxis, and is found by setting x = 0 in the equation and solving for y. Similarly, the xintercept is the xcoordinate of the point where the graph crosses the xaxis, and is found by setting y = 0 and solving for x. Math Page of 9 Section.
5 Equations of Lines There are four forms of an equation of a line: PointSlope Form: y y m( x x ) where m is the slope and (, ) =, x y is a point on the line. SlopeIntercept Form: y = mx + b, where m is the slope and b is the yintercept of the line. Standard Form: Ax + By = C, where A, B, and C real numbers and are written as integers whenever possible*, and A and B cannot both be equal to zero. General Form: Ax + By + C = 0, where A, B, and C are real numbers and are written as integers whenever possible*, and A and B cannot both be equal to zero. *Notes about standard and general form: Standard form and general form are to be written such that A, B, and C are integers whenever possible. In this course, you will be given problems where it is always possible to change the equation so that A, B, and C are integers. There are cases where it is not possible to change the coefficients to integers, such as the equation x + π y = 7, but such examples will not be used in this course. The equations for standard and general form are not unique, as seen in the example below. However, textbooks often display standard and general form so that A > 0, and so that A, B, and C are relatively prime. Example : Change each of the following equations to slopeintercept form, standard form, and general form. 7 A. y = ( x 6) B. x 6 = 8y C. y x = Solution: A. To change to slopeintercept form, y = mx + b, we want to distribute the solve for y. and then 7 y = ( x 6), so 7 0 y = x Math Page of 9 Section.
6 0 y = x +, so 7 7 Slopeintercept form is y = x + + = x + + = x y = x We want to clear all fractions in order to put the equation in standard or general form. We can do this by multiplying by 7 (the common denominator of the fractions): 8 7( y) = 7 x + 7 7, so 7 y = x + 8. We can then rearrange the terms to obtain standard and general form. The equation x + 7 y = 8 is in standard form. The equation x + 7y 8 = 0 is in general form. Note that standard form and general form are not unique. Other acceptable answers can be obtained by choosing any nonzero integer and then multiplying each term in the equation by that integer. B. To change to slopeintercept form, y = mx + b, we can divide both sides of the equation by 8 to solve for y. x 6 = 8y, so x 6 6 y = = x = x Slopeintercept form is y = x. Since the original equation already has terms with integer coefficients, we can simply rearrange the terms of x 6 = 8y to obtain standard and general form. The equation x 8y = 6 is in standard form, and the equation x 8y 6 = 0 is in general form. Although both of the previous answers are acceptable, notice that the first term is negative, and that all three terms have a common factor of. We can divide each term by to obtain x + y = for standard form and x + y + = 0 for general form. These are the answers that would most commonly be given in a textbook answer key. Note that standard form and general form are not unique. Other acceptable answers can be obtained by choosing any nonzero integer and then multiplying each term in the equation by that integer. C. Since the equation y x = 7 contains multiple fractions, we will choose to clear the 9 6 equation of fractions by multiplying each term by the common denominator of 9,, and 6 which is 6. Math Page 6 of 9 Section.
7 7 6 y x 6 6 y 6 x 6 7 =, so =. Simplifying each term, we obtain y 6 x 6 7 =, which gives us the equation 6y x =. 9 6 To change to slopeintercept form, y = mx + b, we need to solve for y: 6y x =, so 6y = x +. Dividing by 6, x + y = = x To change 6y x = to standard form and general form, we can simply rearrange the terms, since the coefficients and the constant term are already integers. The equation x + 6y = is in standard form, and the equation x + 6y = 0 is in general form. Although both of the previous equations are acceptable, it is common to make the first term positive by multiplying each term by, to obtain a standard form of x 6y =, and a general form of x 6y + = 0. Note that standard form and general form are not unique. Other acceptable answers can be obtained by choosing any nonzero integer and then multiplying each term in the equation by that integer. Example : Write an equation in slopeintercept form for the line shown below. Solution: Slopeintercept form is y = mx + b, where m is the slope and b is the yintercept. We can see from the graph that the yintercept is, so b =. We now need to find the slope, m. Choose any two points on the graph with integer coordinates; we will use ( 0, ) and From the point ( 0, ), we move down units, then to the right unit to get to (, 0 )., 0. Math Page 7 of 9 Section.
8 rise So m = = =. We can now substitute m = and b = into y = mx + b and obtain the run equation of the line: y = x + Example : Write an equation in slopeintercept form for the line shown below. Solution: Slopeintercept form is y = mx + b, where m is the slope and b is the yintercept. We can see from the graph that the yintercept is 0, so b = 0. We now need to find the slope, m. Choose any two points on the graph with integer coordinates; we will use ( 0, 0 ) and From the point ( 0, 0 ), we move up units, then to the right units to get to (, ).,. Math Page 8 of 9 Section.
9 So rise m = =. We can now substitute run equation of the line: y = x m = and b = 0 into y = mx + b and obtain the Example 6: Write an equation in slopeintercept form for the line shown below. Solution: Slopeintercept form is y = mx + b, where m is the slope and b is the yintercept. Notice on the graph that the yintercept is not an integer; we will use other means to find that later. Let us first find the slope, m. Choose any two points on the graph with integer coordinates; we will use (, ) and (, ). From the point (, ) right units to get to (, )., we move down units, then to the So m =. We can now use the slope along with one of the two chosen points and substitute into either y mx b = + or y y m( x x ) = to find the equation of the line. (There are four means of solving this problem, using either of the two points with either of the two equations. All methods yield the same result. We will show two of the four methods below.) First, let us plug the point (, ) and the known slope,, into the equation y = mx + b. Math Page 9 of 9 Section.
10 y = mx + b = ( ) + b 6 = + b b = = = Now that we have found the slope and the yintercept, we can substitute y = mx + b to obtain the equation of the line. 0 y = x m = and 0 b = into If we instead use the point (, ) with the pointslope formula, we obtain the same result: y y = m x x y ( ) = ( x ( ) ) y + = ( x + ) Next, distribute the y + = x 6 y = x = x 0 y = x This equation and solve for y. 0 y = x is in slopeintercept form, and matches the answer we found using the first method. Notice that when using y y m( x x ) =, we did not need to separately find the value of b. After solving for y, the equation is in slopeintercept form. Math Page 0 of 9 Section.
11 Example 7: Write an equation in slopeintercept form for the line shown below. Solution: Slopeintercept form is y = mx + b, where m is the slope and b is the yintercept. Once again, notice that the yintercept is not an integer. Let us first find the slope, m. Choose any two points on the graph with integer coordinates; we will choose (, ) and (, ), we move up units, then to the right units to get to (, ).,. From the point So m =. Next, we can use the slope and either of the two chosen points and substitute into = + or y y m( x x ) either y mx b =, to find the equation of the line. Let us use the point (, ) and the known slope,. Two methods are shown, but any one method is sufficient. y = mx + b y y = m x x = ( ) + b y ( ) = ( x ( ) ) 9 9 = + b y + = x b = + = + = y = x + = x + y = mx + b = x + y = x + Math Page of 9 Section.
12 The equation of the line is y = x +. Horizontal and Vertical Lines The equation of any horizontal line is of the form y = b, where b is the yintercept of the line. The line therefore passes through the point ( 0, b ). The slope of any horizontal line is zero. The equation of any vertical line is of the form x = a, where a is the xintercept of the line. The line therefore passes through the point ( a, 0). The slope of any vertical line is undefined. Example 8: Write an equation for the line shown below. Solution: The slope of any horizontal line is zero. The form of any horizontal line is y Hence, the equation of the line is y =. Example 9: Write an equation for the line shown below. = b. Math Page of 9 Section.
13 Solution: The slope of any vertical line is undefined. The form of any vertical line is x = a. Hence, the equation of the line is x =. Example 0: Find an equation of the line in slopeintercept form that passes through ( 0, 7) and has slope. Solution: We are given the yintercept and the slope, so we can substitute into the slopeintercept form and obtain the equation of the line: y = x + 7 Example : A line passes through the point ( 0, 8) and has slope. Find an equation of the line in pointslope form. Then write the equation in slopeintercept form. Solution: Let ( x, y ) represent the point ( 0, 8) information into the pointslope equation. y y = m x x y = y + 8 = 0 ( 8) ( x 0) ( x ). It is given that m =. Substitute this The above equation is in pointslope form. We will now solve for y to write the equation in slopeintercept form. y + 8 = 0 y + 8 = x + y = x + 8 y = x ( x ) Pointslope form is y 8 ( x 0) + =, and slopeintercept form is y = x. Math Page of 9 Section.
14 Example : Find an equation of the line in slopeintercept form that passes through the points (, 9) and (, ). Solution: We can find the slope of the line by applying the slope formula. y y m x x = = = 9 8 We can now use the slope and either one of the two given points to write the equation of the line. Two methods are shown below, but one method is sufficient. We will use the point (, 9). y = mx + b y y = m x x = ( ) + b y 9 = ( x ( ) ) 8 9 = + b y 9 = x b = 9 = = y = x + 9 = x y = mx + b = x y = x The equation of the line is 8 y = x. Example : Find an equation of the line in standard form that has xintercept and yintercept 0.7. Solution: We are given the points (, 0 ) and ( 0, 0.7) m = = 0. 0, respectively. First, find the slope. Now use the slope and the given yintercept to write the equation in slopeintercept form. y = 0.x 0.7 The above equation can alternatively be written with fractions: 7 y = x, so y = x 0 We can clear the fractions by multiplying by 0 (the common denominator of and 0). Math Page of 9 Section.
15 y = x 0 0( y) = 0 x 0 0y = x We then rearrange the terms to obtain x + 0y =, which is in standard form, but is not unique. If we want to make the first term positive (which is common), we can multiply each term by to obtain x 0y =. Other acceptable answers can be obtained by choosing any nonzero integer and then multiplying each term in the equation by that integer. Example : Find an equation of the line that passes through ( 9,.) and ( 0.,. ). Solution: First, we use the slope formula to find the slope of the line. y y m x x = = = = Recall that if the slope of a line is zero, the line is a horizontal line. Hence, we have a horizontal line, and its equation is y =.. Example : Find an equation of the line that passes through, 7 and, 7. Solution: First, we need to find the slope of the line. y y m x x = = =, which is undefined Recall that if the slope of a line is undefined, the line is a vertical line. Hence, we have a vertical line and its equation is x =. 7 Parallel and Perpendicular Lines Two nonvertical lines are parallel if and only if their slopes are the same. (Any two vertical lines are parallel to each other, but have undefined slope.) Math Page of 9 Section.
16 Two lines are perpendicular if and only if their slopes are negative reciprocals of each other. The exception to this rule is when one line is vertical (with undefined slope) and the other line is horizontal (with slope zero). These lines are clearly perpendicular to each other, but are not negative reciprocals since we cannot take the reciprocal of undefined. Note that two numbers c and d are negative reciprocals of each other if d =, which means that c d =. c Example 6: Find the negative reciprocal of each of the following numbers. A. B. C. 0. D. 0.7 Solution: A. The negative reciprocal of is. B. The negative reciprocal of is. C. 0. can be written as, which is 00 D. 0.7 can be written as 0, which is 7. Its negative reciprocal is Its negative reciprocal is. Example 7: Find an equation of the line in slopeintercept form that passes through the point, and is parallel to the line y = x. Solution: Notice that the line y = x is in slopeintercept form y = mx + b, and has a slope of. Our desired line is parallel to y = x, and therefore also has a slope of. We can use the slope,, and the given point,,, to find the equation of the line. Two different methods are shown below which yield the same result. Math Page 6 of 9 Section.
17 y = mx + b y y = m x x = + b y = x = + b y = x + b = 7 y = x + 7 y = mx + b = x + 7 The equation of the line in slopeintercept form is y = x + 7. Example 8: Find an equation of the line in general form that passes through the point (, ) and is perpendicular to the line x + y = 0. Solution: To find the slope of the line x + y = 0, we first need to write it in slopeintercept form. x + y = 0 y = x + 0 y = x + The given line, y = x +, has a slope of. Since we want to write the equation of a line perpendicular to the line y = x +, the desired line has a slope equal to the negative reciprocal of, which is. We can now use the slope,,, to find the equation of the desired line. Two different methods are shown below, which yield the same result., along with the given point, b y x b y ( x ) y = mx + b y y = m x x = + = = = + b = 8 y + = x 8 b = 9 y = x 9 y = mx + b = x 9 Using either method above, we obtain the equation y = x 9. Math Page 7 of 9 Section.
18 We want to write the equation in general form, Ax + By + C = 0, where A, B, and C are integers. We can rearrange the terms so that they are all on one side of the equation. x + y + 9 = 0 Remember that when writing in general form, the answers are not unique. Other acceptable answers can be obtained by choosing any nonzero integer and then multiplying each term in the equation by that integer. Example 9: Find an equation of the line in standard form that has xintercept 8 and is perpendicular to the line that passes through (, 7) and ( 6, 0). Solution: We first use the slope formula to find the slope of the line that passes through (, 7) and ( 6, 0). 0 7 = = = 6 y y m x x To find the slope of the desired perpendicular line, we find the negative reciprocal of is, which. We know that the line has xintercept 8, which means that the graph passes through the point ( 8, 0). We can now use the slope,, along with the given point, ( 8, 0 ), to find the equation of the desired line. Two different methods are shown below, which yield the same result. y = mx + b y y = m x x 0 = = = + b y = x + 6 b = b y x 6 y = mx + b = x + Using either method above, we obtain the equation 6 y = x +. Math Page 8 of 9 Section.
19 We want to write the equation in standard form, Ax + By = C, where A, B, and C are integers. We can clear the fractions by multiplying each term by. 6 ( y) = x + y = x + 6 We then rearrange the terms to obtain x + y = 6, which is in standard form, but is not unique. If we want to make the first term positive (which is common), we can multiply each term by to obtain x y = 6. Other acceptable answers can be obtained by choosing any nonzero integer and then multiplying each term in the equation by that integer. Example 0: Find an equation of the line in slopeintercept form that passes through (, ) and is parallel to the line that passes through (, 0) and ( 0, ). Solution: We first use the slope formula to find the slope of the line that passes through (, 0) and ( 0, ). y y m x x = = = = 0 0 Since the equation of the line we wish to write is parallel to the line that passes through (, 0) and ( 0, ), we need to use this same slope, to find, along with the point (, ) the equation of the line. Two different methods are shown below, which yield the same result. y = mx + b y y = m x x = + = = + b y = x + b = + = 8 y = x + + y = mx + b = x + 8 y = x + 8 b y ( x ) The equation of the line is y = x + 8. Math Page 9 of 9 Section.
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