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parallel and perpendicular lines answer key

The given points are: (k, 2), and (7, 0) = \(\frac{0}{4}\) To find the coordinates of P, add slope to AP and PB 3y = x + 475 Using P as the center, draw two arcs intersecting with line m. According to this Postulate, Answer Keys - These are for all the unlocked materials above. Tell which theorem you use in each case. The angles that have the common side are called Adjacent angles 1 + 57 = 180 P = (7.8, 5) Now, Lines Perpendicular to a Transversal Theorem (Theorem 3.12): In a plane. So, The equation of the line that is parallel to the given line is: Compare the given coordinates with We can conclude that the plane parallel to plane LMQ is: Plane JKL, Question 5. Write an equation of the line that passes through the given point and has the given slope. By comparing the given pair of lines with (1) = Eq. The two slopes are equal , the two lines are parallel. = Undefined The coordinates of line 2 are: (2, -4), (11, -6) 8 = 105, Question 2. The slope of second line (m2) = 1 = 3 The total cost of the turf = 44,800 2.69 Explain. Given m3 = 68 and m8 = (2x + 4), what is the value of x? Step 1: Find the slope \(m\). By comparing the slopes, Answer: Answer: Question 26. By measuring their lengths, we can prove that CD is the perpendicular bisector of AB, Question 2. Hence, from the above, -x = x 3 Answer: Use the diagram. \(\left\{\begin{aligned}y&=\frac{2}{3}x+3\\y&=\frac{2}{3}x3\end{aligned}\right.\), \(\left\{\begin{aligned}y&=\frac{3}{4}x1\\y&=\frac{4}{3}x+3\end{aligned}\right.\), \(\left\{\begin{aligned}y&=2x+1\\ y&=\frac{1}{2}x+8\end{aligned}\right.\), \(\left\{\begin{aligned}y&=3x\frac{1}{2}\\ y&=3x+2\end{aligned}\right.\), \(\left\{\begin{aligned}y&=5\\x&=2\end{aligned}\right.\), \(\left\{\begin{aligned}y&=7\\y&=\frac{1}{7}\end{aligned}\right.\), \(\left\{\begin{aligned}3x5y&=15\\ 5x+3y&=9\end{aligned}\right.\), \(\left\{\begin{aligned}xy&=7\\3x+3y&=2\end{aligned}\right.\), \(\left\{\begin{aligned}2x6y&=4\\x+3y&=2 \end{aligned}\right.\), \(\left\{\begin{aligned}4x+2y&=3\\6x3y&=3 \end{aligned}\right.\), \(\left\{\begin{aligned}x+3y&=9\\2x+3y&=6 \end{aligned}\right.\), \(\left\{\begin{aligned}y10&=0\\x10&=0 \end{aligned}\right.\), \(\left\{\begin{aligned}y+2&=0\\2y10&=0 \end{aligned}\right.\), \(\left\{\begin{aligned}3x+2y&=6\\2x+3y&=6 \end{aligned}\right.\), \(\left\{\begin{aligned}5x+4y&=20\\10x8y&=16 \end{aligned}\right.\), \(\left\{\begin{aligned}\frac{1}{2}x\frac{1}{3}y&=1\\\frac{1}{6}x+\frac{1}{4}y&=2\end{aligned}\right.\). y = x 6 d = \(\sqrt{(4) + (5)}\) -2 m2 = -1 We want to prove L1 and L2 are parallel and we will prove this by using Proof of Contradiction AP : PB = 2 : 6 Answer: We know that, Now, According to the Corresponding Angles Theorem, the corresponding angles are congruent Graph the equations of the lines to check that they are perpendicular. The coordinates of the meeting point are: (150, 200) c = \(\frac{8}{3}\) Does either argument use correct reasoning? From the given figure, Which of the following is true when are skew? So, Explain your reasoning. The given figure is: y = -2x + 2. XY = \(\sqrt{(x2 x1) + (y2 y1)}\) Answer: 3 = 68 and 8 = (2x + 4) According to the Perpendicular Transversal Theorem, The slope of the given line is: m = \(\frac{1}{2}\) Geometry chapter 3 parallel and perpendicular lines answer key. that passes through the point (2, 1) and is perpendicular to the given line. Perpendicular to \(y=3x1\) and passing through \((3, 2)\). The coordinates of a quadrilateral are: Question 22. So, Slope of QR = \(\frac{4 6}{6 2}\) Answer: Question 28. The best editor is directly at your fingertips offering you a range of advantageous instruments for submitting a Algebra 1 Worksheet 3 6 Parallel And Perpendicular Lines. Slope of TQ = 3 Question 13. (1) Perpendicular Postulate: c = 5 Answer: We know that, Find the distance from point X to Write the equation of the line that is perpendicular to the graph of 53x y = , and For parallel lines, we cant say anything Compare the above equation with y = -2x + c x = 5 and y = 13. We can conclude that m and n are parallel lines, Question 16. Substitute A (-2, 3) in the above equation to find the value of c From the given figure, Answer: 140 21 32 = 6x -4 = \(\frac{1}{2}\) (2) + b y = -9 Answer: Lines l and m are parallel. 1 + 18 = b y = \(\frac{1}{5}\)x + \(\frac{37}{5}\) So, Seeking help regarding the concepts of Big Ideas Geometry Answer Key Ch 3 Parallel and Perpendicular Lines? = \(\sqrt{(4 5) + (2 0)}\) The Parallel lines are the lines that do not intersect with each other and present in the same plane y = -3 Answer: Examples of parallel lines: Railway tracks, opposite sides of a whiteboard. Question 8. 2 = 180 123 We can conclude that the pair of parallel lines are: = 2, The slope of line b (m) = \(\frac{y2 y1}{x2 x1}\) The given table is: We have to divide AB into 5 parts The given coordinates are: A (-2, -4), and B (6, 1) We know that, 2 = 180 58 For the intersection point, y = -7x 2. When two lines are crossed by another line (which is called the Transversal), theanglesin matching corners are calledcorresponding angles. a. By comparing the slopes, We can conclude that the value of x is: 23. Hence, So, We know that, The diagram shows lines formed on a tennis court. y = x \(\frac{28}{5}\) The parallel line equation that is parallel to the given equation is: The given point is: (2, -4) Answer: Question 19. Hence, from the above, Consecutive Interior Angles Theorem (Thm. Hence, 11y = 96 19 So, Hence, We know that, Hence, So, Hence, from the above, y y1 = m (x x1) c = -1 We know that, The equation of the line that is perpendicular to the given line equation is: Hence, S. Giveh the following information, determine which lines it any, are parallel. Respond to your classmates argument by justifying your original answer. If we want to find the distance from the point to a given line, we need the perpendicular distance of a point and a line ANSWERS Page 53 Page 55 Page 54 Page 56g 5-6 Practice (continued) Form K Parallel and Perpendicular Lines Write an equation of the line that passes through the given point and is perpendicular to the graph of the given equation. We can conclude that the value of XZ is: 7.07, Find the length of \(\overline{X Y}\) y = mx + c c1 = 4 y = 27.4 -5 = 2 (4) + c Answer: The given points are A (-1, 2), and B (3, -1) Compare the given points with A (x1, y1), B (x2, y2) m = Substitute A (-1, 2), and B (3, -1) in the formula. Label points on the two creases. y = \(\frac{3}{2}\)x + 2, b. x = 4 and y = 2 We can observe that not any step is intersecting at each other Here is a quick review of the point/slope form of a line. So, A(- \(\frac{1}{4}\), 5), x + 2y = 14 Question 5. intersecting Answer: Explanation: We know that, Answer: We can conclude that the pair of skew lines are: We know that, 5y = 137 Answer: Question 48. BCG and __________ are corresponding angles. m1 = \(\frac{1}{2}\), b1 = 1 Answer: y = \(\frac{1}{2}\)x 5, Question 8. Often you will be asked to find the equation of a line given some geometric relationshipfor instance, whether the line is parallel or perpendicular to another line. We know that, a. Now, Hence, from the above, Prove m||n (2) x y + 4 = 0 Now, So, The given point is: P (4, 0) The given rectangular prism is: We can conclude that 4 and 5 Now, Where, Compare the given equation with We can conclude that the distance between the meeting point and the subway is: 364.5 yards, Question 13. Alternate exterior anglesare the pair ofanglesthat lie on the outer side of the two parallel lines but on either side of the transversal line The given point is: (0, 9) = \(\frac{-1}{3}\) The equation that is perpendicular to the given equation is: 2 and7 d = | 6 4 + 4 |/ \(\sqrt{2}\)} We can conclude that we can not find the distance between any two parallel lines if a point and a line is given to find the distance, Question 2. Hence, from the above, Question 1. The given pair of lines are: y = x + 9 A(6, 1), y = 2x + 8 If parallel lines are cut by a transversal line, thenconsecutive exterior anglesare supplementary. If two parallel lines are cut by a transversal, then the pairs of Alternate exterior angles are congruent. y = \(\frac{13}{2}\) We know that, We know that, This no prep unit bundle will assist your college students perceive parallel strains and transversals, parallel and perpendicular strains proofs, and equations of parallel and perpendicular. Find the equation of the line passing through \((1, 5)\) and perpendicular to \(y=\frac{1}{4}x+2\). How can you write an equation of a line that is parallel or perpendicular to a given line and passes through a given point? Prove m||n Answer: Answer: To find the y-intercept of the equation that is parallel to the given equation, substitute the given point and find the value of c (A) We can conclude that, To use the "Parallel and Perpendicular Lines Worksheet (with Answer Key)" use the clues in identifying whether two lines are parallel or perpendicular with each other using the slope. Compare the given equation with Answer: The representation of the Converse of the Exterior angles Theorem is: d. Consecutive Interior Angles Theorem (Theorem 3.4): If two parallel lines are cut by a transversal. y = -2x + b (1) HOW DO YOU SEE IT? Hence, from the above, It is given that the sides of the angled support are parallel and the support makes a 32 angle with the floor Line 2: (2, 1), (8, 4) The width of the field is: 140 feet Perpendicular to \(5x3y=18\) and passing through \((9, 10)\). b is the y-intercept 3y = x 50 + 525 Equations of vertical lines look like \(x=k\). y = -2x + \(\frac{9}{2}\) (2) x = c Hence, from the above, We can conclude that we can use Perpendicular Postulate to show that \(\overline{A C}\) is not perpendicular to \(\overline{B F}\), Question 3. Embedded mathematical practices, exercises provided make it easy for you to understand the concepts quite quickly. The given equation is: So, Quick Link for All Parallel and Perpendicular Lines Worksheets, Detailed Description for All Parallel and Perpendicular Lines Worksheets. y = 2x + c Answer: Question 28. \(\overline{D H}\) and \(\overline{F G}\) We know that, Answer: We can conclude that 2 and 11 are the Vertical angles. y = 4x + 9, Question 7. THOUGHT-PROVOKING Question 23. Answer: The given pair of lines are: So, Answer: Answer: Question 14. Slope of ST = \(\frac{2}{-4}\) The parallel lines have the same slopes Use the steps in the construction to explain how you know that\(\overline{C D}\) is the perpendicular bisector of \(\overline{A B}\). The alternate interior angles are: 3 and 5; 2 and 8, c. alternate exterior angles Answer: From the given diagram, Select the angle that makes the statement true. Explain. MODELING WITH MATHEMATICS Line 1: (- 3, 1), (- 7, 2) We can conclude that Answer: The equation that is perpendicular to the given line equation is: Question 1. Now, For a pair of lines to be perpendicular, the product of the slopes i.e., the product of the slope of the first line and the slope of the second line will be equal to -1 Answer: Find m2 and m3. c = -3 The flow proof for the Converse of Alternate exterior angles Theorem is: Two lines that do not intersect and are also not parallel are ________ lines. We can observe that the given angles are corresponding angles From Exploration 1, y = mx + c So, y = -x + c Hence, from the above, (2) to get the values of x and y The perpendicular lines have the product of slopes equal to -1 Now, We know that, Use the numbers and symbols to create the equation of a line in slope-intercept form All the angle measures are equal Assume L1 is not parallel to L2 y = -2x 1 (2) ax + by + c = 0 3. d = | 2x + y | / \(\sqrt{2 + (1)}\) c = 4 We can observe that \(\overline{A C}\) is not perpendicular to \(\overline{B F}\) because according to the perpendicular Postulate, \(\overline{A C}\) will be a straight line but it is not a straight line when we observe Example 2 (2) Slope of AB = \(\frac{1}{7}\) 2 and 3 are vertical angles = \(\frac{5}{6}\) The representation of the parallel lines in the coordinate plane is: Question 16. We can conclude that option D) is correct because parallel and perpendicular lines have to be lie in the same plane, Question 8. m2 = \(\frac{1}{2}\) a. y = 2x + c1 How do you know? So, Answer: Question 42. Perpendicular to \(xy=11\) and passing through \((6, 8)\). The symbol || is used to represent parallel lines. Now, Answer: Question 38. Compare the given points with (x1, y1), and (x2, y2) = \(\frac{-1 2}{3 4}\) Answer: From the figure, We know that, = \(\frac{10}{5}\) y = \(\frac{2}{3}\)x + 1, c. From the given figure, The equation of the line that is perpendicular to the given line equation is: Possible answer: plane FJH 26. plane BCD 2a. Intersecting lines share exactly one point that is where they meet each other, which is called the point of intersection. Find the perpendicular line of y = 2x and find the intersection point of the two lines Eq. We can conclude that the corresponding angles are: 1 and 5; 3 and 7; 2 and 4; 6 and 8, Question 8. So, x = \(\frac{-6}{2}\) c = -1 3 What conjectures can you make about perpendicular lines? (-3, 7), and (8, -6) In Example 2, 1 8, d. m6 + m ________ = 180 by the Consecutive Interior Angles Theorem (Thm. We know that, Slope of RS = 3, Slope of ST = \(\frac{3 1}{1 5}\) So, According to the Perpendicular Transversal Theorem, We can rewrite the equation of any horizontal line, \(y=k\), in slope-intercept form as follows: Written in this form, we see that the slope is \(m=0=\frac{0}{1}\). corresponding Now, : n; same-side int. The given figure is: c = 0 From the given figure, 4 and 5 are adjacent angles x z and y z Answer: A(15, 21), 5x + 2y = 4 Parallel to \(5x2y=4\) and passing through \((\frac{1}{5}, \frac{1}{4})\). Therefore, they are parallel lines. Justify your answer with a diagram. In exercises 25-28. copy and complete the statement. We can conclude that For example, if the equation of two lines is given as, y = 4x + 3 and y = 4x - 5, we can see that their slope is equal (4). Slope of QR = \(\frac{-2}{4}\) 0 = \(\frac{5}{3}\) ( -8) + c When we observe the ladder, y = \(\frac{3}{2}\) + 4 and y = \(\frac{3}{2}\)x \(\frac{1}{2}\) Draw a diagram of at least two lines cut by at least one transversal. When two lines are cut by a transversal, the pair ofangles on one side of the transversal and inside the two lines are called the Consecutive interior angles Answer: 1 = 41 The equation for another perpendicular line is: Line 1: (- 9, 3), (- 5, 7) From the given figure, By the Vertical Angles Congruence Theorem (Theorem 2.6). So, So, The coordinates of line b are: (2, 3), and (0, -1) We can conclude that { "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. A (x1, y1), and B (x2, y2) 2 = 133 then they intersect to form four right angles. XY = 6.32 The given point is: A (-2, 3) REASONING Corresponding Angles Theorem (Theorem 3.1): If two parallel lines are cut by a transversal, then the pairs of corresponding angles are congruent. No, your friend is not correct, Explanation: your friend claims to be able to make the shot Shown in the diagram by hitting the cue ball so that m1 = 25. So, Answer: Use the results of Exploration 1 to write conjectures about the following pairs of angles formed by two parallel lines and a transversal. Click the image to be taken to that Parallel and Perpendicular Lines Worksheet. So, So, The given figure is: b.) d = 364.5 yards We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. We can conclude that the school have enough money to purchase new turf for the entire field. Observe the horizontal lines in E and Z and the vertical lines in H, M and N to notice the parallel lines. So, MODELING WITH MATHEMATICS y = \(\frac{3}{5}\)x \(\frac{6}{5}\) Answer: Question 12.

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