We can observe that 2x = 7 By the _______ . The given figure is: Now, -1 = \(\frac{1}{2}\) ( 6) + c y = 3x + 2 m2 = \(\frac{1}{2}\), b2 = -1 y = 3x 6, Question 20. Question 5. y= 2x 3 ERROR ANALYSIS The equation of the line along with y-intercept is: The parallel lines have the same slopes In a plane, if a line is perpendicular to one of the two parallel lines, then it is perpendicular to the other line also = \(\frac{1}{3}\), The slope of line c (m) = \(\frac{y2 y1}{x2 x1}\) (5y 21) = (6x + 32) = \(\frac{8 0}{1 + 7}\) These worksheets will produce 6 problems per page. Answer: Question 15. In this form, you can see that the slope is \(m=2=\frac{2}{1}\), and thus \(m_{}=\frac{1}{2}=+\frac{1}{2}\). We can conclude that the line that is parallel to the given line equation is: Slope (m) = \(\frac{y2 y1}{x2 x1}\) Answer: Question 26. Slope of AB = \(\frac{-4 2}{5 + 3}\) 2 = 180 123 When we compare the converses we obtained from the given statement and the actual converse, We can conclude that the slope of the given line is: 3, Question 3. The Parallel lines are the lines that do not intersect with each other and present in the same plane Answer: Identify the slope and the y-intercept of the line. We know that, (x1, y1), (x2, y2) The equation for another line is: Now, From the given figure, m1 and m3 Substitute (0, -2) in the above equation The given equation is: Hence, from the above, If two parallel lines are cut by a transversal, then the pairs of consecutive interior angles formed are supplementary Answer: Question 24. So, Now, corresponding The given figure is: Answer: By using the Alternate exterior angles Theorem, x z and y z Each unit in the coordinate plane corresponds to 10 feet y = 0.66 feet So, Explain your reasoning. (1) and eq. Now, (a) parallel to and Now, You are designing a box like the one shown. The given point is: A (3, -1) In Exercises 21 and 22, write and solve a system of linear equations to find the values of x and y. If the slopes of the opposite sides of the quadrilateral are equal, then it is called as Parallelogram Using the same compass selling, draw an arc with center B on each side \(\overline{A B}\). x y + 4 = 0 The coordinates of the school = (400, 300) The give pair of lines are: -3 = -4 + c Answer: The line through (k, 2) and (7, 0) is perpendicular to the line y = x \(\frac{28}{5}\). The given line equation is: Hence, from the above, Write the equation of a line that would be parallel to this one, and pass through the point (-2, 6). We can conclude that the distance between the given 2 points is: 17.02, Question 44. y = \(\frac{1}{3}\)x + c Now, We can conclude that we can use Perpendicular Postulate to show that \(\overline{A C}\) is not perpendicular to \(\overline{B F}\), Question 3. So, 8 6 = b Also the two lines are horizontal e. m1 = ( 7 - 5 ) / ( -2 - (-2) ) m2 = ( 13 - 1 ) / ( 5 - 5 ) The two slopes are both undefined since the denominators in both m1 and m2 are equal to zero. We know that, Write an equation of the line passing through the given point that is perpendicular to the given line. So, The perimeter of the field = 2 ( Length + Width) Answer: y = x + c Answer: We know that, Answer: Question 42. How do you know? ERROR ANALYSIS Answer: Question 22. Explain why the tallest bar is parallel to the shortest bar. c = 1 The given figure is: We know that, So, Answer: It is given that If two parallel lines are cut by a transversal, then the pairs of alternate exterior angles are congruent The slopes of the parallel lines are the same Substitute (1, -2) in the above equation We have to find 4, 5, and 8 We have to find the distance between X and Y i.e., XY All perpendicular lines can be termed as intersecting lines, but all intersecting lines cannot be called perpendicular because they need to intersect at right angles. The coordinates of line 1 are: (10, 5), (-8, 9) COMPLETE THE SENTENCE It is given that We can observe that Now, Answer: perpendicular, or neither. Question 11. Slope of RS = \(\frac{-3}{-1}\) = 0 Is your classmate correct? Answer: Now, Hence, from the given figure, Hence, from the above, The converse of the Alternate Interior angles Theorem: The given equation is: m1m2 = -1 So, The slope of the equation that is parallel t the given equation is: \(\frac{1}{3}\) Draw a diagram to represent the converse. The slopes are equal fot the parallel lines The equation of the line that is parallel to the line that represents the train tracks is: Parallel lines are always equidistant from each other. The given point is:A (6, -1) We can observe that the given angles are the corresponding angles Answer: Answer: y = 3x 5 We know that, = \(\frac{6}{2}\) The given point is: A (-2, 3) 4.5 Equations of Parallel and Perpendicular Lines Solving word questions We know that, Prove the statement: If two lines are horizontal, then they are parallel. The line that passes through point F that appear skew to \(\overline{E H}\) is: \(\overline{F C}\), Question 2. So, Answer: F if two coplanar strains are perpendicular to the identical line then the 2 strains are. y = \(\frac{2}{3}\)x + b (1) In the equation form of a line y = mx +b lines that are parallel will have the same value for m. Perpendicular lines will have an m value that is the negative reciprocal of the . BCG and __________ are corresponding angles. EG = \(\sqrt{(5) + (5)}\) 12y = 156 y = \(\frac{1}{3}\)x 2 -(1) Perpendicular to \(y=3x1\) and passing through \((3, 2)\). In a square, there are two pairs of parallel lines and four pairs of perpendicular lines. In which of the following diagrams is \(\overline{A C}\) || \(\overline{B D}\) and \(\overline{A C}\) \(\overline{C D}\)? So, It is given that m || n = \(\frac{-1}{3}\) Write the equation of the line that is perpendicular to the graph of 9y = 4x , and whose y-intercept is (0, 3). The equation for another parallel line is: From the given figure, On the other hand, when two lines intersect each other at an angle of 90, they are known as perpendicular lines. You decide to meet at the intersection of lines q and p. Each unit in the coordinate plane corresponds to 50 yards. y1 = y2 = y3 We know that, We can observe that the pair of angle when \(\overline{A D}\) and \(\overline{B C}\) are parallel is: APB and DPB, b. y = \(\frac{2}{3}\) The points are: (-2, 3), (\(\frac{4}{5}\), \(\frac{13}{5}\)) Answer: y = \(\frac{13}{5}\) Answer: Write the equation of the line that is perpendicular to the graph of 53x y = , and Hence, from the above, We can conclude that Unit 3 (Parallel & Perpendicular Lines) In this unit, you will: Identify parallel and perpendicular lines Identify angle relationships formed by a transversal Solve for missing angles using angle relationships Prove lines are parallel using converse postulate and theorems Determine the slope of parallel and perpendicular lines Write and graph Question 11. y = -x + c In Exercises 21-24. are and parallel? 2 = 133 y = \(\frac{2}{3}\)x + 9, Question 10. m = 2 Substitute A (-2, 3) in the above equation to find the value of c We know that, Parallel and perpendicular lines can be identified on the basis of the following properties: If the slope of two given lines is equal, they are considered to be parallel lines. A(3, 1), y = \(\frac{1}{3}\)x + 10 Question 31. We can conclude that the lines that intersect \(\overline{N Q}\) are: \(\overline{N K}\), \(\overline{N M}\), and \(\overline{Q P}\), c. Which lines are skew to ? The missing information the student assuming from the diagram is: m a, n a, l b, and n b transv. We can conclude that a line equation that is perpendicular to the given line equation is: When we compare the given equation with the obtained equation, We can conclude that the given pair of lines are parallel lines. Question 3. (x1, y1), (x2, y2) The lines that have the slopes product -1 and different y-intercepts are Perpendicular lines Now, So, These worksheets will produce 6 problems per page. Then use a compass and straightedge to construct the perpendicular bisector of \(\overline{A B}\), Question 10. What is m1? So, So, Hence, So, 2x = 135 15 Two lines are cut by a transversal. = 44,800 square feet { "3.01:_Rectangular_Coordinate_System" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "3.02:_Graph_by_Plotting_Points" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "3.03:_Graph_Using_Intercepts" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "3.04:_Graph_Using_the_y-Intercept_and_Slope" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "3.05:_Finding_Linear_Equations" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "3.06:_Parallel_and_Perpendicular_Lines" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "3.07:_Introduction_to_Functions" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "3.08:_Linear_Inequalities_(Two_Variables)" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "3.0E:_3.E:_Review_Exercises_and_Sample_Exam" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()" }, { "00:_Front_Matter" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "01:_Real_Numbers_and_Their_Operations" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "02:_Linear_Equations_and_Inequalities" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "03:_Graphing_Lines" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "04:_Solving_Linear_Systems" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "05:_Polynomials_and_Their_Operations" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "06:_Factoring_and_Solving_by_Factoring" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "07:_Rational_Expressions_and_Equations" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "08:_Radical_Expressions_and_Equations" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "09:_Solving_Quadratic_Equations_and_Graphing_Parabolas" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "10:_Appendix_-_Geometric_Figures" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "zz:_Back_Matter" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()" }, [ "article:topic", "license:ccbyncsa", "showtoc:no", "authorname:anonymous", "licenseversion:30", "program:hidden" ], https://math.libretexts.org/@app/auth/3/login?returnto=https%3A%2F%2Fmath.libretexts.org%2FBookshelves%2FAlgebra%2FBeginning_Algebra%2F03%253A_Graphing_Lines%2F3.06%253A_Parallel_and_Perpendicular_Lines, \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}}}\) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\), Finding Equations of Parallel and Perpendicular Lines, status page at https://status.libretexts.org. The sum of the angle measures of a triangle is: 180 So, The slope of perpendicular lines is: -1 Proof of the Converse of the Consecutive Exterior angles Theorem: y = \(\frac{3}{2}\) Hence, from the above, Slope of line 2 = \(\frac{4 6}{11 2}\) (2) y = -2x 1 (2) Answer: Is your friend correct? In Exercises 13 16. write an equation of the line passing through point P that s parallel to the given line. Now, We can conclude that We can conclude that the given statement is not correct. Write an equation of the line that passes through the point (1, 5) and is Hence, from the above, Perpendicular to \(y3=0\) and passing through \((6, 12)\). b.) Answer: Question 6. The coordinates of P are (7.8, 5). Slope of AB = \(\frac{5}{8}\) c = -2 We know that, x = 4 and y = 2 We know that, Make a conjecture about what the solution(s) can tell you about whether the lines intersect. So, In this form, we can see that the slope of the given line is \(m=\frac{3}{7}\), and thus \(m_{}=\frac{7}{3}\). The slopes of the parallel lines are the same We can conclude that 18 and 23 are the adjacent angles, c. Two nonvertical lines in the same plane, with slopes m1 and m2, are parallel if their slopes are the same, m1 = m2. x = 54 Find the equation of the line passing through \((\frac{7}{2}, 1)\) and parallel to \(2x+14y=7\). We want to prove L1 and L2 are parallel and we will prove this by using Proof of Contradiction ANALYZING RELATIONSHIPS Answer: The coordinates of y are the same. 2 = 2 (-5) + c m2 = -2 Question 5. So, By using the Vertical Angles Theorem, The representation of the given pair of lines in the coordinate plane is: Draw \(\overline{P Z}\), Question 8. Examine the given road map to identify parallel and perpendicular streets. Hence, from the above, So, The given point is: (-1, 5) Let the given points are: WHICH ONE did DOESNT BELONG? The portion of the diagram that you used to answer Exercise 26 on page 130 is: Question 2.
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