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AnswerQuestion Answers for Class 12Class 12 BiologyClass 12 ChemistryClass 12 EnglishClass 12 MathsClass 12 PhysicsClass 12 Social ScienceClass 12 Business StudiesClass 12 EconomicsQuestion Answers for Class 11Class 11 EconomicsClass 11 Computer ScienceClass 11 BiologyClass 11 ChemistryClass 11 EnglishClass 11 MathsClass 11 PhysicsClass 11 Social ScienceClass 11 AccountancyClass 11 Business StudiesQuestion Answers for Class 10Class 10 ScienceClass 10 EnglishClass 10 MathsClass 10 Social ScienceClass 10 General KnowledgeQuestion Answers for Class 9Class 9 General KnowledgeClass 9 ScienceClass 9 EnglishClass 9 MathsClass 9 Social ScienceQuestion Answers for Class 8Class 8 ScienceClass 8 EnglishClass 8 MathsClass 8 Social ScienceQuestion Answers for Class 7Class 7 ScienceClass 7 EnglishClass 7 MathsClass 7 Social ScienceQuestion Answers for Class 6Class 6 ScienceClass 6 EnglishClass 6 MathsClass 6 Social ScienceQuestion Answers for Class 5Class 5 ScienceClass 5 EnglishClass 5 MathsClass 5 Social ScienceQuestion Answers for Class 4Class 4 ScienceClass 4 EnglishClass 4 MathsFind the equation of the perpendicular bisector of the line segment joining points (7, 1) and (3, 5).AnswerVerified600.9k+ viewsHint: Perpendicular bisector is a perpendicular line which passes through the midpoint of the line joining the given two points. Start with finding the midpoint of the given points and then use this point to find the equation of line perpendicular to the line formed by the given two points.Complete step-by-step answer:Let the given points (7, 1) and (3, 5) be A and B respectively.Also suppose point M is the midpoint of points A (7, 1) and B (3, 5). And we know that the coordinates of midpoint of two points $\left( {{x_1},{y_1}} \right)$ and $\left( {{x_2},{y_2}} \right)$ is $\left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}} \right)$.So the coordinates of points M are:\[\therefore \]Midpoint of $\left( {AB} \right) = \left( {\dfrac{{7 + 3}}{2},\dfrac{{5 + 1}}{2}} \right) = \left( {\dfrac{{10}}{2},\dfrac{6}{2}} \right) = M\left( {5,3} \right)$ Further, we know that the slope of line joining two points $\left( {{x_1},{y_1}} \right)$ and $\left( {{x_2},{y_2}} \right)$ is $\dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$.Therefore, the slope of the line joining points A and B is ${m_1} = \dfrac{{3 - 7}}{{5 - 1}} = \dfrac{{ - 4}}{4} = - 1$ Perpendicular bisector of points A and B will be the line perpendicular to AB. Since the product of slopes of two perpendicular lines is -1, we have:$ \Rightarrow {m_1} \times {m_2} = - 1$, where ${m_2}$ is the slope of the perpendicular bisector. $ \Rightarrow - 1 \times {m_2} = - 1$$ \Rightarrow {m_2} = 1$Perpendicular bisector will pass through the points A and B i.e. point M.In this case, the perpendicular bisector is eventually a line passing through point $M\left( {5,3} \right)$ and having slope ${m_2} = 1$.According to point slope form of equation of line, if a line of slope $m$ is passing through a point $\left( {{x_1},{y_1}} \right)$, its equation is:$ \Rightarrow \left( {y - {y_1}} \right) = m\left( {x - {x_1}} \right)$So, using point slope form of the equation of line, the equation of perpendicular bisector is:$ \Rightarrow (y - 3) = 1(x - 5)$$ \Rightarrow y - 3 = x - 5$$ \Rightarrow x - y - 2 = 0$ Thus the equation of the perpendicular bisector is $x - y - 2 = 0$.Note: In the above problem, we have used point slope form to find the equation of line.If the line is passing through two known points $\left( {{x_1},{y_1}} \right)$ and $\left( {{x_2},{y_2}} \right)$, then its equation is:$ \Rightarrow \left( {y - {y_1}} \right) = \left( {\dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}} \right)\left( {x - {x_1}} \right)$There are various other ways to find the equation of line. We can use different methods as per the data available to us.Recently Updated PagesMaster Class 10 General Knowledge: Engaging Questions & Answers for SuccessMaster Class 10 Science: Engaging Questions & Answers for SuccessMaster Class 10 Social Science: Engaging Questions & Answers for SuccessMaster Class 10 Maths: Engaging Questions & Answers for SuccessMaster Class 10 English: Engaging Questions & Answers for SuccessMaster Class 10 Computer Science: Engaging Questions & Answers for SuccessMaster Class 10 General Knowledge: Engaging Questions & Answers for SuccessMaster Class 10 Science: Engaging Questions & Answers for SuccessMaster Class 10 Social Science: Engaging Questions & Answers for SuccessMaster Class 10 Maths: Engaging Questions & Answers for SuccessMaster Class 10 English: Engaging Questions & Answers for SuccessMaster Class 10 Computer Science: Engaging Questions & Answers for Success

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