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**Equations of Lines Equations of Lines**

Linear Equations Equations of Lines This module delves into the various descriptions of lines and their features. Linear Equations 7/2/2013

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**Lines and Equations Point-Slope Form**

Equations of Lines Point-Slope Form Given line L and point (x1, y1) on L Let (x, y) be any other point on L Find slope y x L (x, y) (x1, y1) ∆y = y – y1 m = ∆y ∆x Point-Slope Form The point-slope form of a linear equation allows us to write the equation with minimal information: the slope and the coordinates of one point on the line. In the illustration, we are given one point, namely (x1, y1), and we want a general equation for the line. We use the definition of the slope and any other arbitrary point (x, y) to form y and x. As soon as we have written the slope as the ratio of these two quantities, we have an equation of the line, in the sense that any point (x, y) on the line represents an ordered pair (x, y) that is a solution of the equation, and conversely. The equation is called the point-slope form because it clearly represents the slope of the line and the coordinates of one point on the line. This equation represents the whole line because we chose (x, y) to be any other point on the line, a way of saying that it represents all other points on the line. ∆x = x – x1 7/2/2013 Linear Equations Linear Equations 7/2/2013

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**Lines and Equations Point-Slope Form m = = y – y1 = m(x –x1)**

Equations of Lines Point-Slope Form m = ∆y ∆x y x = y – y1 x – x1 L (x, y) y – y1 = m(x –x1) (x1, y1) ∆y = y – y1 Point-Slope Form The point-slope form of a linear equation allows us to write the equation with minimal information: the slope and the coordinates of one point on the line. In the illustration, we are given one point, namely (x1, y1), and we want a general equation for the line. We use the definition of the slope and any other arbitrary point (x, y) to form y and x. As soon as we have written the slope as the ratio of these two quantities, we have an equation of the line, in the sense that any point (x, y) on the line represents an ordered pair (x, y) that is a solution of the equation, and conversely. The equation is called the point-slope form because it clearly represents the slope of the line and the coordinates of one point on the line. This equation represents the whole line because we chose (x, y) to be any other point on the line, a way of saying that it represents all other points on the line. The final form shown is sometimes referred to as the modified point-slope form. The original form is simply ∆x = x – x1 Point-Slope Form 7/2/2013 Linear Equations Linear Equations 7/2/2013

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**Lines and Equations Slope-Intercept Form**

Equations of Lines Slope-Intercept Form Consider a non-vertical line L Locate y-intercept Let (x, y) be any other point on the line Find slope m y x L (x, y) Lines and Equations: Slope-Intercept Form Here, given a non-vertical line L we write the equation of the line in general, using the vertical intercept (y-intercept) as the specified point on the line. We don’t yet know both intercept coordinates, so we write the equation in point-slope form using b and 0. Rewriting this as mx = y – b we see that moving b to the other side of the equation produces the standard slope-intercept form mx + b = y . ∆y = y – b (0, b) ∆x = x – 0 7/2/2013 Linear Equations Linear Equations 7/2/2013

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**Lines and Equations Slope-Intercept Form m = = m(x – 0) = y – b**

Equations of Lines Slope-Intercept Form m = ∆y ∆x y x L (0, b) (x, y) ∆x = x – 0 ∆y = y – b = y – b x – 0 m(x – 0) = y – b Solving for y, Lines and Equations: Slope-Intercept Form Here, given a non-vertical line L we write the equation of the line in general, using the vertical intercept (y-intercept) as the specified point on the line. We don’t yet know both intercept coordinates, so we write the equation in point-slope form using b and 0. Rewriting this as mx = y – b we see that moving b to the other side of the equation produces the standard slope-intercept form mx + b = y . y = mx + b Slope-Intercept Form 7/2/2013 Linear Equations Linear Equations 7/2/2013

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**Examples Point-Slope Form A line through (5, 10) with slope**

Equations of Lines Point-Slope Form A line through (5, 10) with slope Find the equation: 4 5 y x m = 4 5 = y – 10 x – 5 m = 4/5 Point-Slope Form Example In this example we are given a specific point and a slope number and asked to find the equation of the line through the given point with the given slope. We can write down the equation immediately in point-slope form, one of the advantages of using this form. This requires no intermediate calculations. This form is not unique, since the slope 4/5 can also be written as 8/10, or 12/15, or …. Any slope that reduces to 4/5 can be used, meaning there are infinitely many ways to write the equation. (5,10) Question: Is this form unique ? 7/2/2013 Linear Equations Linear Equations 7/2/2013

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**Examples Slope-Intercept Form**

Equations of Lines Slope-Intercept Form A line with intercept (0, 4) and slope -3 Find the equation: y = mx + b = -3x + 4 y x Slope-Intercept Form Example In this example, we are again given a specific point and a slope value. In this case we are asked to find the equation of the line, in slope-intercept form, through the point with the given slope. In this case, since the given point is the y-intercept, we can immediately write down the equation as y = mx + b the typical slope-intercept form. This form is unique, since there is only one value that will fit the slope number and the coefficient of y is simply 1. Note that if, in the second example, the given point were not the y-intercept, we would have to do some calculations to arrive at the slope-intercept form. (0, 4) Question: Is this form unique ? m = -3 7/2/2013 Linear Equations Linear Equations 7/2/2013

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**Lines and Equations Standard Form**

Equations of Lines Standard Form Algebraic form not directly graph related Useful for systems of linear equations For constants A, B and C, with B ≠ 0 Ax + By = C Can we still find slope and intercepts ? Standard Form Any linear equation can written in standard form, so equations in either point-slope form or slope-intercept form can be converted to standard form. The big advantage for standard form becomes apparent when dealing with systems of linear equations, especially those with more than two variables. This form is then adaptable to the matrix techniques of linear algebra. Question: Is this form unique ? 7/2/2013 Linear Equations Linear Equations 7/2/2013

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**Lines and Equations a ≠ A Standard Form – Slope m**

Equations of Lines Standard Form – Slope Rewriting the equation: This has slope-intercept form with y = x A B – + C Note: a ≠ A = mx + b b ≠ B , Standard Form Any linear equation can written in standard form, so equations in either point-slope form or slope-intercept form can be converted to standard form. The big advantage for standard form becomes apparent when dealing with systems of linear equations, especially those with more than two variables. This form is then adaptable to the matrix techniques of linear algebra. Note that the slope and the coordinates of both intercepts can be read directly from the coefficients in the equation. The form is not unique, since all the coefficients can be multiplied by any constant k to produce an equivalent equation. If B = 0 then the equation becomes Ax = C or x = C/B which means x is constant and the graph is a horizontal line. Also note the constants B and b are do not represent the same number. As in all of mathematics, upper and lower case characters are generally used to represent different things. Slope = A B – = m 7/2/2013 Linear Equations Linear Equations 7/2/2013

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**( ) Lines and Equations Standard Form – Intercepts m**

Equations of Lines Standard Form Rewriting the equation: – Intercepts y = x A B – + C = mx + b So … = m A B – Note: b ≠ B C B b = Standard Form Any linear equation can written in standard form, so equations in either point-slope form or slope-intercept form can be converted to standard form. The big advantage for standard form becomes apparent when dealing with systems of linear equations, especially those with more than two variables. This form is then adaptable to the matrix techniques of linear algebra. Note that the slope and the coordinates of both intercepts can be read directly from the coefficients in the equation. The form is not unique, since all the coefficients can be multiplied by any constant k to produce an equivalent equation. If B = 0 then the equation becomes Ax = C or x = C/B which means x is constant and the graph is a horizontal line. Also note the constants B and b are do not represent the same number. As in all of mathematics, upper and lower case characters are generally used to represent different things. = b m – C B A ( ) = C A Fractions 7/2/2013 Linear Equations Linear Equations 7/2/2013

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**( ) ( ) ( ) Lines and Equations Standard Form – Intercepts a ≠ A**

Equations of Lines Standard Form Rewriting the equation: Intercepts – Intercepts y = x A B – + C Note: a ≠ A ( ) 0, C B = y-intercept = (0, b) b ≠ B Standard Form Any linear equation can written in standard form, so equations in either point-slope form or slope-intercept form can be converted to standard form. The big advantage for standard form becomes apparent when dealing with systems of linear equations, especially those with more than two variables. This form is then adaptable to the matrix techniques of linear algebra. Note that the slope and the coordinates of both intercepts can be read directly from the coefficients in the equation. The form is not unique, since all the coefficients can be multiplied by any constant k to produce an equivalent equation. If B = 0 then the equation becomes Ax = C or x = C/B which means x is constant and the graph is a horizontal line. Also note the constants B and b are do not represent the same number. As in all of mathematics, upper and lower case characters are generally used to represent different things. ( ) = b m – , = ( ) C A , 0 (a, 0) = x-intercept 7/2/2013 Linear Equations Linear Equations 7/2/2013

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**Standard Form Example Intercepts with Standard Form**

Equations of Lines Intercepts with Standard Form Consider equation 6x + 5y = 30 When x = 0 , 5y = 30 y = 6 Vertical intercept is (0, 6) x y (0, 6) Finding Intercepts from Standard Form Given the equation in standard form, it is a simple matter to find the intercepts by letting first one variable, and then the other, be zero. As shown, this easily establishes both intercepts. By simply remembering that intercepts always occur on one axis or the other, we know that one of the variables is zero. By setting each variable to 0 and solving the remaining equation, we immediately see the coordinates of each of the intercepts. This reinforces the concept of the intercept as a point, not a number. 7/2/2013 Linear Equations Linear Equations 7/2/2013

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**Standard Form Example Intercepts with Intercept Form**

Equations of Lines Intercepts with Intercept Form Consider equation 6x + 5y = 30 When y = 0, 6x = 30 x = 5 Horizontal intercept is (5, 0) x y (0, 6) (5, 0) Finding Intercepts from Standard Form Given the equation in standard form, it is a simple matter to find the intercepts by letting first one variable, and then the other, be zero. As shown, this easily establishes both intercepts. By simply remembering that intercepts always occur on one axis or the other, we know that one of the variables is zero. By setting each variable to 0 and solving the remaining equation, we immediately see the coordinates of each of the intercepts. This reinforces the concept of the intercept as a point, not a number. 7/2/2013 Linear Equations Linear Equations 7/2/2013

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**( ) Lines and Equations Intercept Form +**

Equations of Lines Intercept Form Consider standard form equation Ax + By = C For C ≠ 0 , this becomes A C x + B y = 1 Finding Intercepts from the Intercept Form Given the equation in standard form, it is a simple matter to find convert it to intercept form by dividing by C, assuming of course the C ≠ 0. This yields and by multiplying and dividing each term by its coefficient we can rewrite as which is known as the intercept form. It is easy to see that the fraction dividing x is the x-coordinate of the x-intercept by just letting y = 0 and solving for x. Similarly, letting x = 0 and solving for y yields the y-coordinate of the y-intercept. Thus, the intercepts are where a and b are the usual x and y coordinates for the intercepts. The above calculations assume that C ≠ 0. But what if C = 0? Then either the line crosses both axes at the origin, that is, the intercepts are the same point, or the line lies on one of the axes (x-axis if A = 0, or y-axis if B = 0). OR intercept form = 1 ( ) C A x + y B Fractions 7/2/2013 Linear Equations Linear Equations 7/2/2013

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**( ) ) ( ( ) Lines and Equations Intercept Form + + a**

Equations of Lines Intercept Form Consider intercept form equation ( ) C A x y B + = 1 a x + b y = 1 x y Note: A ≠ a 0, C B ) ( ( ) 0, b = Finding Intercepts from the Intercept Form Given the equation in standard form, it is a simple matter to find convert it to intercept form by dividing by C, assuming of course the C ≠ 0. This yields and by multiplying and dividing each term by its coefficient we can rewrite as which is known as the intercept form. It is easy to see that the fraction dividing x is the x-coordinate of the x-intercept by just letting y = 0 and solving for x. Similarly, letting x = 0 and solving for y yields the y-coordinate of the y-intercept. Thus, the intercepts are where a and b are the usual x and y coordinates for the intercepts. The above calculations assume that C ≠ 0. But what if C = 0? Then either the line crosses both axes at the origin, that is, the intercepts are the same point, or the line lies on one of the axes (x-axis if A = 0, or y-axis if B = 0). B ≠ b , 0 C A ( ) Question: (a, ) = What if C = 0 ? 7/2/2013 Linear Equations Linear Equations 7/2/2013

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**Standard Form Example Finding Intercepts with Intercept Form ( )**

Equations of Lines Finding Intercepts with Intercept Form Consider equation 6x + 5y = 30 Vertical intercept is (0, 6) Horizontal intercept is (5, 0) x y (0, 6) 30 1 6x + 5y ( ) ( ) = x 5 y 6 + = 1 (5, 0) Finding Intercepts from Standard Form Given the equation in standard form, it is a simple matter to find the intercepts by letting first one variable, and then the other, be zero. As shown, this easily establishes both intercepts. By simply remembering that intercepts always occur on one axis or the other, we know that one of the variables is zero. By setting each variable to 0 and solving the remaining equation, we immediately see the coordinates of each of the intercepts. This reinforces the concept of the intercept as a point, not a number. 7/2/2013 Linear Equations Linear Equations 7/2/2013

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**Horizontal and Vertical Lines**

Equations of Lines Horizontal Lines Form: y = k for some constant k From standard form Ax + By = C when A = 0, B ≠ 0 C B y = Horizontal Lines Here we show why the slope of a horizontal line is 0. It should be emphasized that this does NOT mean there is no slope. The slope of 0 is a perfectly good slope number. The lines y = 0 and x = 0 are in fact the x-axis and y-axis, respectively. Question: What is the line y = 0 called ? The x-axis ! 7/2/2013 Linear Equations Linear Equations 7/2/2013

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**Horizontal and Vertical Lines**

Equations of Lines Horizontal Lines Example: y = 3 Pick any points (x1, 3) , (x2, 3) Slope m is then Note: Zero slope is NOT the same as no slope x y (x1, 3) (x2, 3) y = 0 y = 3 m = y x 3 – 3 x2 – x1 = x1 x2 Horizontal Lines Here we show why the slope of a horizontal line is 0. It should be emphasized that this does NOT mean there is no slope. The slope of 0 is a perfectly good slope number. The lines y = 0 and x = 0 are in fact the x-axis and y-axis, respectively. x = x2 – x1 7/2/2013 Linear Equations Linear Equations 7/2/2013

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**Horizontal and Vertical Lines**

Equations of Lines Vertical Lines Form: x = k for some constant k From standard form Ax + By = C when A ≠ 0, B = 0 Then Ax = C Vertical Lines Here again we attempt to compute the slope of the vertical line and find we must divide by 0 to do it. Since division by 0 is not allowed in our number system, we cannot compute a slope for a vertical line. This line truly has no slope. One might ask why we are not allowed to divide by 0. The answer is not that we aren’t intelligent enough, but that we must assign some number to the result. It is easily shown that any number we might assign, including 0 itself, leads to one or more contradictions. That is, we can then prove something to be true that is already known (and proved) to be false. Hence, as a matter of mathematical philosophy, we choose not to divide by 0. A C x = Question: What is the line x = 0 ? The y-axis ! 7/2/2013 Linear Equations Linear Equations 7/2/2013

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**Horizontal and Vertical Lines**

Equations of Lines Vertical Line Example: x = 4 Pick any points (4, y1) , (4, y2) Slope m is then x y x = 4 y2 (4, y2) y y1 (4, y1) m = y x = 4 – 4 y2 – y1 x = 0 Vertical Lines Here again we attempt to compute the slope of the vertical line and find we must divide by 0 to do it. Since division by 0 is not allowed in our number system, we cannot compute a slope for a vertical line. This line truly has no slope. One might ask why we are not allowed to divide by 0. The answer is not that we aren’t intelligent enough, but that we must assign some number to the result. It is easily shown that any number we might assign, including 0 itself, leads to one or more contradictions. That is, we can then prove something to be true that is already known (and proved) to be false. Hence, as a matter of mathematical philosophy, we choose not to divide by 0. = y2 – y1 Undefined ! Note: No slope is not the same as zero slope ! 7/2/2013 Linear Equations Linear Equations 7/2/2013

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**Parallel Lines Parallel Lines Horizontal Lines**

Equations of Lines Parallel Lines Horizontal Lines Zero slope, always parallel Vertical Lines No slope, always parallel Other lines Lines with same slope, always parallel Parallel Lines We note that the key feature of parallel lines is the same slope, or in the case of vertical lines, no slope. In the illustration we find the equation of a line, given a point on the line and the equation of a parallel line. The parallel line gives us the slope of the line we must find. With the given point, we find the equation of the line just as we did in finding the equation of a line given its slope and a point on the line. 7/2/2013 Linear Equations Linear Equations 7/2/2013

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**Parallel Lines Parallel Lines Example**

Equations of Lines Parallel Lines Example Find the equation of the line through (4, 10) parallel to the line Slope of new line is Point-slope form is y 1 2 – = x + 6 1 2 – x y (4, 10) = y – 10 x – 4 1 2 – Parallel Lines We note that the key feature of parallel lines is the same slope, or in the case of vertical lines, no slope. In the illustration we find the equation of a line, given a point on the line and the equation of a parallel line. The parallel line gives us the slope of the line we must find. With the given point, we find the equation of the line just as we did in finding the equation of a line given its slope and a point on the line. y = – (½)x + 12 y = – (½)x + 6 OR = y – 10 (x – 4) 1 2 – 7/2/2013 Linear Equations Linear Equations 7/2/2013

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**Perpendicular Lines in General**

Equations of Lines Step 1 Geometry gives us x y L1 m1 = b1 a , b1 = am1 m2 c1 b1 am1 = c12 = a2 + b12 a b2 – am2 = c2 m2 = – b2 a Perpendicular Lines in General The relationship between perpendicular lines is a little more complex than that between parallel lines. Given perpendicular lines L1 and L2 with slopes m1 and m2, respectively, as shown in the illustration. Pick a point on line L1 above the intersection point and drop a vertical line from this new point to the x-axis. Project a horizontal line through the intersection of L1 and L2 as shown. We can now form three distinct triangles. The horizontal segment marked a forms a leg of two distinct triangles, as shown. We apply the Pythagorean Theorem to the upper triangle, with leg b1 lying on the vertical projection line and hypotenuse c1 lying on line L1, giving c12 = a2 + b12 The lower triangle is similarly formed using the same leg a, leg b2 on the vertical projection line, and the hypotenuse c2 on line L2 . This gives us c22 = a2 + b22 The third triangle, with legs c1 and c2 and hypotenuse b1 + b2 give us c12 + c22 = (b1 + b2)2 Combining this with the first two equations we have c12 + c22 = a2 + b12 + a2 + b22 = (b1 + b2)2 leading to b12 + 2b1b2 + b22 = b12 + 2a2 + b22 b2 = – am2 , m1 L2 c22 = a2 + b22 c12 + c22 = b12 + 2a2 + b22 7/2/2013 Linear Equations Linear Equations 7/2/2013

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**Perpendicular Lines in General**

Equations of Lines Step 2 x y c12 + c22 = b12 + b22 + 2a2 L1 c12 + c22 = (b1 + b2)2 m2 c1 = b12 + 2b1b2 + b22 b1 b1 + b2 a 2a2 = 2b1b2 b2 c2 = 2(am1)(– am2) Perpendicular Lines in General The relationship between perpendicular lines is a little more complex than that between parallel lines. Given perpendicular lines L1 and L2 with slopes m1 and m2, respectively, as shown in the illustration. Pick a point on line L1 above the intersection point and drop a vertical line from this new point to the x-axis. Project a horizontal line through the intersection of L1 and L2 as shown. We can now form three distinct triangles. The horizontal segment marked a forms a leg of two distinct triangles, as shown. We apply the Pythagorean Theorem to the upper triangle, with leg b1 lying on the vertical projection line and hypotenuse c1 lying on line L1, giving c12 = a2 + b12 The lower triangle is similarly formed using the same leg a, leg b2 on the vertical projection line, and the hypotenuse c2 on line L2 . This gives us c22 = a2 + b22 The third triangle, with legs c1 and c2 and hypotenuse b1 + b2 give us c12 + c22 = (b1 + b2)2 Combining this with the first two equations we have c12 + c22 = a2 + b12 + a2 + b22 = (b1 + b2)2 leading to b12 + 2b1b2 + b22 = b12 + 2a2 + b22 m1 L2 = – 2a2m1m2 1 = – m1m2 – 1 m2 m1 = 7/2/2013 Linear Equations Linear Equations 7/2/2013

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**Perpendicular Lines Example**

Equations of Lines Example Find the equation of the line through (2, 3) perpendicular to line y = –(⅓)x + 3 Slope of the given line is m1 = –⅓ Slope of the new line is x y (2, 3) y = –(⅓)x + 3 Perpendicular Lines – Example We use the slope relationship we just derived for perpendicular lines, namely m2 = – We are given the line y = (–½)x + 6 we want to find the equation of the line through point (10,10) perpendicular to the given line. We know that the slopes have the negative reciprocal relationship shown above. Since the given line has slope –½ then the perpendicular line has slope 2. This line goes through point (10,10) has a point-slope equation y – 10 = 2(x – 10) which can be rewritten in other forms, such as y = 2x – 10 or in “standard form” this is 2x – y = –10 . m1 1 1 m1 – m2 = = – 1 ⅓ = 3 7/2/2013 Linear Equations Linear Equations 7/2/2013

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**Perpendicular Lines Example**

Equations of Lines Example Slope of the new line is x y (2, 3) y = –(⅓)x + 3 = 3 m2 1 m1 – ⅓ y = 3x – 3 Alternate point-slope form y – y1 = m(x – x1) y – 3 = 3(x – 2) Perpendicular Lines – Example We use the slope relationship we just derived for perpendicular lines, namely m2 = – We are given the line y = (–⅓)x + 3 and want to find the equation of the line through point (2,3) perpendicular to the given line. We know that the slopes have the negative reciprocal relationship shown above. Since the given line has slope –⅓ then the perpendicular line has slope 3. This line goes through point (2,3) has a point-slope equation y – 3 = 3(x – 2) which can be rewritten in other forms, such as y = 3x – 3 or in “standard form” this is 3x – y = 3 . Slope-intercept form m1 1 y = mx – mx1 + y1 y = 3x – 3 7/2/2013 Linear Equations Linear Equations 7/2/2013

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**Variation Direct Variation**

Equations of Lines Direct Variation A variable y varies directly as variable x if y = kx for some constant k The constant k is called the constant of variation K is also known as the constant of proportionality Direct Variation The way a dependent variable changes with changes in the independent variable is known as variation. If y = kx for some constant k, then y varies directly with x. That is, if x increases, so does y. And if x decreases, so does y. If y = k/x for some constant k, then y varies inversely with x. That, if x increases, then y decreases. And if y decreases, then x increases. The constant k is known as the constant of variation or constant of proportionality. The direct variation relationship, y = kx, is represented as a linear equation whose graph has a zero intercept (i.e. zero initial value) and slope k. 7/2/2013 Linear Equations Linear Equations 7/2/2013

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**Variation Direct Variation Example k**

Equations of Lines Direct Variation Example State sales tax t varies directly as the amount of sale s , i.e. t = ks For tax of $200 on a $12.50 sale, what is the constant of variation ? k = t s s t Direct Variation The way a dependent variable changes with changes in the independent variable is known as variation. If y = kx for some constant k, then y varies directly with x. That is, if x increases, so does y. And if x decreases, so does y. If y = k/x for some constant k, then y varies inversely with x. That, if x increases, then y decreases. And if y decreases, then x increases. The constant k is known as the constant of variation or constant of proportionality. The direct variation relationship, y = kx, is represented as a linear equation whose graph has a zero intercept (i.e. zero initial value) and slope k. = 12.50 200.00 k = .0625 .0625 = Question: Does this look like y = mx + b ? 7/2/2013 Linear Equations Linear Equations 7/2/2013

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**Variation Inverse Variation**

Equations of Lines Inverse Variation Variable y varies inversely as variable x if for constant of variation k k is also known as the constant of inverse proportionality y = k x x y Inverse Variation As with direct variation, a dependent variable can change value as the independent variable changes, but in a different way. If y = k/x for some constant k, then y varies inversely with x. That is, if x increases then y decreases. And if x decreases then y increases. The inverse variation relationship, y = k/x, is a non-linear relationship. The graph for this relationship is of course a hyperbola. 7/2/2013 Linear Equations Linear Equations 7/2/2013

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**Variation Inverse Variation Example**

Equations of Lines Inverse Variation Example At constant temperature the pressure P of a gas in a balloon is inversely proportional to its volume V so that V P P = k V Inverse Variation As with direct variation, a dependent variable can change value as the independent variable changes, but in a different way. Since both P and V are positive, then k must also be positive. Theoretically, as the volume approaches zero, the pressure increases without bound (i.e., “goes to infinity”). Conversely, as the volume V increases without bound, the pressure P approaches zero. In the real world, however, there are physical limits on volume and pressure, so that volume of gas cannot go to zero, and pressure cannot become infinite. 7/2/2013 Linear Equations Linear Equations 7/2/2013

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**Think about it ! Equations of Lines Linear Equations 7/2/2013**

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**Perpendicular Lines in General**

Equations of Lines Step 1 Geometry gives us x y L1 m1 = b1 a , b1 = am1 c12 = a2 + b12 m2 c1 b1 am1 = = a2 + (am1)2 a b2 – am2 = c2 m2 = – b2 a b2 = – am2 , Perpendicular Lines in General The relationship between perpendicular lines is a little more complex than that between parallel lines. Given perpendicular lines L1 and L2 with slopes m1 and m2, respectively, as shown in the illustration. Pick a point on line L1 above the intersection point and drop a vertical line from this new point to the x-axis. Project a horizontal line through the intersection of L1 and L2 as shown. We can now form three distinct triangles. The horizontal segment marked a forms a leg of two distinct triangles, as shown. We apply the Pythagorean Theorem to the upper triangle, with leg b1 lying on the vertical projection line and hypotenuse c1 lying on line L1, giving c12 = a2 + b12 The lower triangle is similarly formed using the same leg a, leg b2 on the vertical projection line, and the hypotenuse c2 on line L2 . This gives us c22 = a2 + b22 The third triangle, with legs c1 and c2 and hypotenuse b1 + b2 give us c12 + c22 = (b1 + b2)2 Combining this with the first two equations we have c12 + c22 = a2 + b12 + a2 + b22 = (b1 + b2)2 leading to b12 + 2b1b2 + b22 = b12 + 2a2 + b22 m1 L2 c22 = a2 + b22 = a2 + (am2)2 c12 + c22 = a2 + (am1)2 + a2 + (am2)2 7/2/2013 Linear Equations Linear Equations 7/2/2013

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**Parallel and Perpendicular Lines**

Equations of Lines Perpendicular Lines in General (continued) Thus Subtracting like terms gives Recalling that b1 = am1 and b2 = – am2 gives x y L2 L1 m2 m1 a b1 c1 b2 c2 am1 = – am2 b12 + 2b1b2 + b22 = b12 + 2a2 + b22 2b1b2 = 2a2 2(am1)(– am2) = 2a2 Perpendicular Lines in General Carrying forward the equation b12 + 2b1b2 + b22 = b12 + 2a2 + b22 we subtract like terms from both sides, giving 2b1b2 = 2a2 or simply b1b2 = a2 As shown on the diagram, b1 = am1 and b2 = –am2 , so substituting in the above equation we get 2(am1 )(–am2 ) = 2a2 This simplifies to m1m2 = –1 Or we express as m2 = – The bottom line here is that the slopes of perpendicular lines are the negative reciprocals of each other. – 2a2m1m2 = 2a2 m1m2 = –1 OR m2 = 1 m1 – 7/2/2013 Linear Equations m1 1 Linear Equations 7/2/2013

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**Parallel and Perpendicular Lines**

Equations of Lines Perpendicular Lines in General (continued) Thus Subtracting like terms gives Recalling that b1 = am1 and b2 = – am2 gives x y L2 L1 m2 m1 a b1 c1 b2 c2 am1 = – am2 b12 + 2b1b2 + b22 = b12 + 2a2 + b22 2b1b2 = 2a2 2(am1)(– am2) = 2a2 Perpendicular Lines in General Carrying forward the equation b12 + 2b1b2 + b22 = b12 + 2a2 + b22 we subtract like terms from both sides, giving 2b1b2 = 2a2 or simply b1b2 = a2 As shown on the diagram, b1 = am1 and b2 = –am2 , so substituting in the above equation we get 2(am1 )(–am2 ) = 2a2 This simplifies to m1m2 = –1 Or we express as m2 = – The bottom line here is that the slopes of perpendicular lines are the negative reciprocals of each other. – 2a2m1m2 = 2a2 m1m2 = –1 OR m2 = 1 m1 – 7/2/2013 Linear Equations m1 1 Linear Equations 7/2/2013

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Algebra 2 Chapter 2.

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