Substitute A (3, -4) in the above equation to find the value of c x = \(\frac{108}{2}\) d = | 2x + y | / \(\sqrt{2 + (1)}\) Is your friend correct? ATTENDING TO PRECISION = 2 Department of Education (1) = Eq. What is the perimeter of the field? So, Hence, from the above, a. Using P as the center and any radius, draw arcs intersecting m and label those intersections as X and Y. Hence, Answer: WRITING Answer: Question 34. In Exploration 2, Now, Now, Hence, from the given figure, Answer: 2x y = 18 We can observe that the pair of angle when \(\overline{A D}\) and \(\overline{B C}\) are parallel is: APB and DPB, b. Prove: t l So, A new road is being constructed parallel to the train tracks through points V. An equation of the line representing the train tracks is y = 2x. The product of the slopes is -1 Prove c||d m = \(\frac{5}{3}\) Now, Then, select Insert statistic chart > histogram > choose Pareto. The representation of the given pair of lines in the coordinate plane is: y = mx + b Justify your answer with a diagram. Hence, from the above, It is given that You can find all the concepts via the quick links available below. We can conclude that the distance from the given point to the given line is: \(\frac{4}{5}\). 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 Label the point of intersection as Z. Answer: THOUGHT-PROVOKING XY = \(\sqrt{(3 + 3) + (3 1)}\) Slope of AB = \(\frac{4}{6}\) (A) PY Hence, from the above, In geometry, A triangle is shape whose three sides are all the same y = -2x + 1 WebHome; Math; Geometry; Triangle area calculator - step by step calculation, formula & solved example problem to find the area for the given values of base b, & height h of triangle in different measurement units between inches (in), feet (ft), meters (m), centimeters (cm) & millimeters (mm). So, We can conclude that the converse we obtained from the given statement is true Select the angle that makes the statement true. Now, BCG and __________ are consecutive interior angles. By comparing the slopes, We can observe that, d = | -2 + 6 |/ \(\sqrt{5}\) The given figure is: The given figure is: From Exploration 2, Answer: Question 26. Two or more objects or points are said to be equidistantif they are at the same distance from a place. Hence, from the above, Name two pairs of supplementary angles when \(\overline{A B}\) and \(\overline{D C}\) are parallel. Slope of TQ = 3 The given equation is: It is given that m || n According to the Vertical Angles Theorem, the vertical angles are congruent c = 0 So, Hence, from the above, Let us learn how to make use of the Pareto chart for various applications. Two lines, a and b, are perpendicular to line c. Line d is parallel to line c. The distance between lines a and b is x meters. Now, c = 2 + 2 Now, Hence, -2 = 3 (1) + c Question 4. We can conclude that 1 = 60. a is both perpendicular to b and c and b is parallel to c, Question 20. Select the correct answer and click on the Finish buttonCheck your score and answers at the end of the quiz, Visit BYJUS for all JEE related queries and study materials, \(\begin{array}{l}\frac{x}{a} + \frac{y}{b}=1\end{array} \), \(\begin{array}{l}x\cos \alpha +y\sin \alpha =p\end{array} \), \(\begin{array}{l}\frac{x}{p\sec \alpha }+\frac{y}{p\cos ec\alpha }=1\end{array} \), \(\begin{array}{l}x\cos \alpha +y\sin \alpha =P\end{array} \), \(\begin{array}{l}y-{{y}_{1}} = \left( \frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}} \right)\left( x-{{x}_{1}} \right)\end{array} \), \(\begin{array}{l}\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{{{b}_{1}}}{{{b}_{2}}} \neq \frac{{{c}_{1}}}{{{c}_{2}}}\end{array} \), \(\begin{array}{l}\frac{{{a}_{1}}}{{{a}_{2}}} \neq \frac{{{b}_{1}}}{{{b}_{2}}}\end{array} \), \(\begin{array}{l}\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{{{b}_{1}}}{{{b}_{2}}}=\frac{{{c}_{1}}}{{{c}_{2}}}\end{array} \), \(\begin{array}{l}Let\,\,\,\,{{L}_{1}}\,\,\,\,\equiv \,\,\,y={{m}_{1}}x+{{c}_{1}}\end{array} \), \(\begin{array}{l}{{L}_{2}}\,\,\,\equiv \,\,\,y={{m}_{2}}x+{{c}_{2}}\end{array} \), \(\begin{array}{l}\text{Angle} = \theta ={{\tan }^{-1}}\left| \left( \frac{{{m}_{2}}-{{m}_{1}}}{1+{{m}_{1}}{{m}_{2}}} \right) \right|\end{array} \), \(\begin{array}{l}\Rightarrow {{m}_{2}}={{m}_{1}}\,\,\,\,\,\,\,\to \,\,\,\,\,lines\,are\,parallel\\\end{array} \), \(\begin{array}{l}\Rightarrow \,\,{{m}_{1}}{{m}_{2}}=-1,\,\,\,\,\,\,\,\,\,\,lines\,L1\,\And L2\,are\,perpendicular\,to\,each\,other\end{array} \), \(\begin{array}{l}\ell =\left| \frac{a{{x}_{1}}+b{{y}_{1}}+c}{\sqrt{{{a}^{2}}+{{b}^{2}}}} \right|\end{array} \), \(\begin{array}{l}\frac{x-{{x}_{1}}}{a}=\frac{y-{{y}_{1}}}{b}=\frac{-(a{{x}_{1}}+b{{y}_{1}}+c)}{\left( {{a}^{2}}+{{b}^{2}} \right)}\end{array} \), \(\begin{array}{l}\frac{h-{{x}_{1}}}{a}=\frac{k-{{y}_{1}}}{b}=\frac{-2(a{{x}_{1}}+b{{y}_{1}}+c)}{\left( {{a}^{2}}+{{b}^{2}} \right)}\end{array} \), \(\begin{array}{l}\frac{{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}}{\sqrt{{{a}_{1}}^{2}+{{b}_{1}}^{2}}}=\pm \frac{{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}}{\sqrt{{{a}_{2}}^{2}+{{b}_{2}}^{2}}}\end{array} \), \(\begin{array}{l}{{L}_{1}}\equiv {{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0\end{array} \), \(\begin{array}{l}{{L}_{2}}\equiv {{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0\end{array} \), \(\begin{array}{l}{{L}_{3}}\equiv {{a}_{3}}x+{{b}_{3}}y+{{c}_{3}}=0\end{array} \), \(\begin{array}{l}\left| \begin{matrix} {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\ {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\ \end{matrix} \right|=0\end{array} \), \(\begin{array}{l}x=\frac{-(2g+2hy)\pm \sqrt{{{(2g+2hy)}^{2}}-4a(b{{y}^{2}}+2fy+c)}}{2a}\end{array} \), \(\begin{array}{l}x=\frac{-(2g+2hy)\pm \sqrt{Q(y)}}{2a}\end{array} \), \(\begin{array}{l}\left| \begin{matrix} a & h & g \\ h & b & f \\ g & f & c \\ \end{matrix} \right|=0\end{array} \), \(\begin{array}{l}\tan \theta =\left( \left| \frac{2\sqrt{{{h}^{2}}-ab}}{a+b} \right| \right)\end{array} \), \(\begin{array}{l}\Rightarrow {{y}^{2}}+\frac{2h}{b}xy+\frac{ab}{b}{{x}^{2}}=0\end{array} \), \(\begin{array}{l}\Rightarrow m1 + m2 = \frac{2h}{b}\end{array} \), \(\begin{array}{l}m_1 m_2=\frac{a}{b}\end{array} \), \(\begin{array}{l}\tan \theta =\left| \frac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}} \right|=\left| \frac{\sqrt{{{({{m}_{1}}+{{m}_{2}})}^{2}}4{{m}_{1}}{{m}_{2}}}}{1+{{m}_{1}}{{m}_{2}}} \right|=\left| \frac{2\sqrt{{{h}^{2}}-ab}}{a+b} \right|\end{array} \), \(\begin{array}{l}\theta ={{\tan }^{-1}}\left| \left( \frac{{{m}_{2}}-{{m}_{1}}}{1+{{m}_{1}}{{m}_{2}}} \right) \right|\end{array} \), \(\begin{array}{l}\frac{x}{a}+\frac{y}{b}=1\end{array} \), \(\begin{array}{l}P\left( \frac{2a+1.0}{2+1},\frac{2.0+1.b}{2+1} \right)\end{array} \), \(\begin{array}{l}\frac{x}{\left( -15/2 \right)}+\frac{y}{12}=1\end{array} \), \(\begin{array}{l}\Rightarrow \tan \theta =-\sqrt{8}\end{array} \), \(\begin{array}{l}y-2=-\sqrt{8}\left( x-1 \right)\end{array} \), \(\begin{array}{l}\Rightarrow \sqrt{8}x+y-\sqrt{8}-2=0.\end{array} \), \(\begin{array}{l}y-{{y}_{1}}=\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\left( x-{{x}_{1}} \right)\end{array} \), \(\begin{array}{l}y-3=\frac{-2 3}{4 + 1}\left( x+1 \right)\Rightarrow x+y-2=0\end{array} \), \(\begin{array}{l}\frac{7-3}{4-1}=\frac{4}{3}=\tan \alpha\end{array} \), \(\begin{array}{l}-\frac{3}{-4}=\frac{3}{4}=\tan \beta\end{array} \), \(\begin{array}{l}\tan 45{}^\circ =\frac{3-m}{1+3m}\end{array} \), \(\begin{array}{l}{{a}^{2}}+2a-3=0\end{array} \), \(\begin{array}{l}\left( a-1 \right)\left( a+3 \right)=0\end{array} \), \(\begin{array}{l}\Rightarrow \frac{1-\left( -1 \right)}{2-x}=\frac{5-1}{4-2}\end{array} \), \(\begin{array}{l}\Rightarrow \frac{2}{2-x}=2\Rightarrow x=1\end{array} \), \(\begin{array}{l}\tan \theta =\left| \frac{{{m}_{2}}-{{m}_{1}}}{1+{{m}_{1}}{{m}_{2}}} \right|\end{array} \), \(\begin{array}{l}\therefore \frac{1}{3}=\left| \frac{m-2m}{1+\left( 2m \right).m} \right|\Rightarrow \frac{1}{3}=\left| \frac{-m}{1+2{{m}^{2}}} \right|\end{array} \), \(\begin{array}{l}\Rightarrow 2{{\left| m \right|}^{2}}-3\left| m \right|+1=0\end{array} \), \(\begin{array}{l}\Rightarrow \left( \left| m \right|-1 \right)\left( 2\left| m \right|-1 \right)=0\end{array} \), \(\begin{array}{l}\Rightarrow \left| m \right|=1\,\,or\,\,\left| m \right|=1/2\end{array} \), \(\begin{array}{l}\Rightarrow \left| m \right|\pm 1\,\,or\,\,m=\pm 1/2\\\end{array} \), \(\begin{array}{l}{{m}_{1}}=\frac{b-3}{a+1}\end{array} \), \(\begin{array}{l}\therefore \left( \frac{b-3}{a+1} \right)\times \left( \frac{3}{4} \right)=-1\end{array} \), \(\begin{array}{l}a=\frac{68}{25}\end{array} \), \(\begin{array}{l}b=-\frac{49}{25}\end{array} \), \(\begin{array}{l}\frac{\left| 7+3 \right|}{\sqrt{5}}=2\sqrt{5}.\end{array} \), \(\begin{array}{l}\frac{\left| k-\left( -4 \right) \right|}{\sqrt{5}}=2\sqrt{5}\Rightarrow \frac{k+4}{\sqrt{5}}=\pm 2\sqrt{5}\Rightarrow k=6,-14.\end{array} \), \(\begin{array}{l}\frac{4x+3y-6}{\sqrt{{{4}^{2}}+{{3}^{2}}}}=\pm \frac{5x+12y+9}{\sqrt{{{5}^{2}}+{{12}^{2}}}}\end{array} \), \(\begin{array}{l}\tan \theta =\left| \frac{-\frac{4}{3}-\frac{9}{7}}{1+\left( \frac{-4}{3} \right)\frac{9}{7}} \right|=\frac{11}{3}>1.\end{array} \), \(\begin{array}{l}4x+3y-6=4\times 1+3\times 2-6>0,\end{array} \), \(\begin{array}{l}5x+12y+9=5\times 1+12\times 2+9>0.\end{array} \), \(\begin{array}{l}\frac{4x+3y-6}{5}=\frac{5x+12y+9}{13}\Rightarrow 9x-7y-41=0.\end{array} \), \(\begin{array}{l}6\lambda +2\left( 7 \right)\left( 4 \right)\left( \frac{7}{2} \right)-2{{\left( 7 \right)}^{2}}-3{{\left( 4 \right)}^{2}}-\lambda {{\left( \frac{7}{2} \right)}^{2}}=0\end{array} \), \(\begin{array}{l}\Rightarrow 6\lambda +196-98-48-\frac{49\lambda }{4}=0\end{array} \), \(\begin{array}{l}\Rightarrow \frac{49\lambda }{4}-6\lambda =196-146=50\end{array} \), \(\begin{array}{l}\Rightarrow \frac{25\lambda }{4}=50\,\,\\\lambda =\frac{200}{25}=8\end{array} \), \(\begin{array}{l}{{x}^{2}}\left( a+2h+b \right)=0\end{array} \), \(\begin{array}{l}\tan \theta =\frac{\pm 2\sqrt{\frac{25}{4}-6}}{2+3}\\ \theta ={{\tan }^{-1}}\left|( \frac{1}{5} \right)|\end{array} \), \(\begin{array}{l}\left( \sqrt{3}x-y \right)\left( x-\sqrt{3}y \right)=0.\end{array} \), \(\begin{array}{l}x\left( \sqrt{3}x-y \right)=0\end{array} \), \(\begin{array}{l}\sqrt{3}{{x}^{2}}-xy=0\end{array} \), A line is a geometry object characterized under zero width object that extends on both sides. The given figure is: Understanding Artificial Intelligence - Major concepts for enterprise applica Four Public Speaking Tips From Standup Comedians, How to Fortify a Diverse Workforce to Battle the Great Resignation, Six Business Lessons From 10 Years Of Fantasy Football, Irresistible content for immovable prospects, How To Build Amazing Products Through Customer Feedback. The given pair of lines are: In Exercises 3 6. find the coordinates of point P along the directed line segment AB so that AP to PB is the given ratio. If the product of slopes of two lines in the plane is $-1$, then the lines are perpendicular and vice-versa. So, Explain your reasoning. reviewed by educators from public and private schools, colleges, and/or m = \(\frac{1}{6}\) and c = -8 Hence, from the above figure, Explain your reasoning. d = | x y + 4 | / \(\sqrt{1 + (-1)}\) Hence, from the above, If two lines are parallel to the same line, then they are parallel to each other 2 = 41 Originally slope of the line is tan = m. Now the slope of the line after rotation is 3. 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 Question 23. Hence, y = -3x + 650, b. \(\frac{5}{2}\)x = 5 We will take as an example that you know the value of the distance of points A and B. REASONING Hence, from the above, The given pair of lines are: This topic introduces you to basic Geometry, primarily focusing on the properties of the angles formed: i) when two lines intersect each other The representation of the given pair of lines in the coordinate plane is: 1 + 138 = 180 Hence, from the above, Is there enough information in the diagram to conclude that m || n? It is given that a coordinate plane has been superimposed on a diagram of the football field where 1 unit is 20 feet. (x1, y1), (x2, y2) We know that, x1 = x2 = x3 . The converse of the given statement is: From the given figure, The product of the slopes of perpendicular lines is equal to -1 x = \(\frac{69}{3}\) a. m5 + m4 = 180 //From the given statement From the given figure, [1] This assumption is used in calculations of business profit or population of any country. Choose an appropriate measurement such as frequency, quantity, cost and time. Hence, We can conclude that the alternate interior angles are: 3 and 6; 4 and 5, Question 7. Compare the given points with (x1, y1), and (x2, y2) The two pairs of perpendicular lines are l and n, c. Identify two pairs of skew line The conjecture about \(\overline{A B}\) and \(\overline{c D}\) is: x = 180 73 So, So, Given angle bisector. We can observe that According to the consecutive Interior Angles Theorem, Substitute (-5, 2) in the above equation y = 3x 5 Answer: Answer: Answer: y = \(\frac{1}{2}\)x + 8, Question 19. We can conclude that m || n by using the Corresponding Angles Theorem, Question 14. So, So, Linear Pair Perpendicular Theorem (Thm. The lines that have the slopes product -1 and different y-intercepts are Perpendicular lines The given equation is: So, So, Hence, from the above, In spherical geometry, all points are points on the surface of a sphere. Graph the equations of the lines to check that they are parallel. Now, Question 27. m = \(\frac{0 2}{7 k}\) A gazebo is being built near a nature trail. Slope of line 1 = \(\frac{-2 1}{-7 + 3}\) 1 = 123 and 2 = 57. w y and z x We have to find the point of intersection Answer: Answer: \(\frac{5}{2}\)x = 2 3x = 69 We can find the equation (by solving first for b) if we have a point and the slope. The given equation is: Answer: Answer: We know that, Hence,f rom the above, Answer: Question 34. Line 2: (7, 0), (3, 6) ATTENDING TO PRECISION = \(\frac{-1 3}{0 2}\) So, Answer: Hence, So, The slopes are equal fot the parallel lines We can conclude that We know that, Answer: Question 12. For a pair of lines to be coincident, the pair of lines have the same slope and the same y-intercept This contradicts what was given,that angles 1 and 2 are congruent. The equation that is perpendicular to the given equation is: Click Calculate to get the slope of line. So, Step 4: In Exercises 11 and 12. prove the theorem. Answer: Slope of AB = \(\frac{4 3}{8 1}\) When we compare the given equation with the obtained equation, We can conclude that a line equation that is perpendicular to the given line equation is: 2 = \(\frac{1}{2}\) (-5) + c Question 38. By using the corresponding angles theorem, In Euclidean geometry, the two perpendicular lines form 4 right angles whereas, In spherical geometry, the two perpendicular lines form 8 right angles according to the Parallel lines Postulate in spherical geometry. The slope of first line (m1) = \(\frac{1}{2}\) = \([(1/2 - (-1))^2 + (5/2 - 2)^2]\) X (-3, 3), Y (3, 1) We can conclude that the number of points of intersection of parallel lines is: 0, a. Answer: We know that, Alternate exterior anglesare the pair ofanglesthat lie on the outer side of the two parallel lines but on either side of the transversal line We can conclude that the consecutive interior angles of BCG are: FCA and BCA. Answer: alternate interior, alternate exterior, or consecutive interior angles. Does either argument use correct reasoning? The coordinates of line d are: (-3, 0), and (0, -1) 2 = 2 (-5) + c So, WebThis nature of change of function is expressed in the sign of the slope. Answer: Answer: In the diagram below. y = mx + c y = \(\frac{24}{2}\) We know that, a. For the analysis of the revenue growth of the organisation with respect to the time period. We can conclude that both converses are the same So, We can observe that a. m1 + m8 = 180 //From the given statement Answer: Answer: k 7 = -2 The slope of the line of the first equation is: Write an equation of the line that is (a) parallel and (b) perpendicular to the line y = 3x + 2 and passes through the point (1, -2). Let us take an example, where we need to prepare a chart of feedback analysis for XYZ restaurant, as per the reviews and ratings received from the customers. We can conclude that y = 2x + 1 In spherical geometry, all points are points on the surface of a sphere. 1 + 2 = 180 A triangle has vertices L(0, 6), M(5, 8). Yes, since the midpoint of a line segment is at an equal distance from the two endpoints, it is said to be equidistant from them. The lines that have the same slope and different y-intercepts are Parallel lines Answer: (13, 1), and (9, -4) Substitute (0, -2) in the above equation Hence, Now, = (\(\frac{-5 + 3}{2}\), \(\frac{-5 + 3}{2}\)) We know that, WebSolution to Problem 3: Use the distance formula to write an equation in x. Answer: From the given figure, 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 y = \(\frac{1}{3}\)x + 10 Substitute (0, 1) in the above equation So, We can conclude that Compare the given points with Obtain the slope of a second line when the slope of first line is 2. universities. From the given coordinate plane, justifies the statement which states that-If two sides of two adjacent acute angles are perpendicular, then the angles are complementary. If you even interchange the second and third statements, you could still prove the theorem as the second line before interchange is not necessary Answer: Hence, Hence,f rom the above, The given point is: P (3, 8) The representation of the given coordinate plane along with parallel lines is: Considering that point B is a bisector of the segment line, conclude that the distance of points B and C is also equal to 20 cm. 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