If (a,0), (0, B) And (p, Q) Are Collinear. Prove - Vedantu

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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 MathsIf (a,0), (0, b) and (p, q) are collinear. Prove:$\dfrac{p}{a} + \dfrac{q}{b} = 1$AnswerVerified579.3k+ viewsHint: We will use the concept that area of a triangle formed by three collinear points is 0. The formula for area of triangle in coordinate geometry is given as:Area of $\Delta = \dfrac{1}{2}\left| {{x_1}({y_2} - {y_3}) + {x_2}({y_3} - {y_1}) + {x_3}({y_1} - {y_2})} \right|$ where, $({x_1},{y_1}),({x_2},{y_2})$ and $({x_3},{y_3})$ are the coordinates of the vertices of the triangle.Complete step by step solution: Now, we are given three points (a, 0), (0, b) and (p, q). Let these three points be the three vertices $A({x_1},{y_1}),B({x_2},{y_2})$and $C({x_3},{y_3})$. then,$ ({x_1},{y_1}) = (a,0) \\ ({x_2},{y_2}) = (0,b) \\ ({x_3},{y_3}) = (p,q) \\ $.Now, area of $\Delta ABC = \dfrac{1}{2}\left| {{x_1}({y_2} - {y_3}) + {x_2}({y_3} - {y_1}) + {x_3}({y_1} - {y_2})} \right|$Now, putting the values of the coordinates of A, B and C we will get the area of $\Delta ABC$:$ \Delta ABC = \dfrac{1}{2}\left| {{x_1}({y_2} - {y_3}) + {x_2}({y_3} - {y_1}) + {x_3}({y_1} - {y_2})} \right| \\\Rightarrow 0 = \dfrac{1}{2}\left| {a(b - q) + 0(q - 0) + p(0 - b)} \right| \\\Rightarrow 0 = \dfrac{1}{2}\left| {ab - aq + 0 - 0 + 0 - pb} \right| \\\Rightarrow 0 = \dfrac{1}{2}\left| {ab - aq - bp} \right| \\\Rightarrow 0 = \left| {ab - aq - bp} \right| \\\Rightarrow ab = aq + bp \\ $Now dividing the both sides by ab, we will get:$ \Rightarrow \dfrac{{ab}}{{ab}} = \dfrac{{aq}}{{ab}} + \dfrac{{bp}}{{ab}} \\\Rightarrow 1 = \dfrac{q}{b} + \dfrac{p}{a} \\\Rightarrow \dfrac{p}{a} + \dfrac{q}{b} = 1 \\ $Hence proved that $\dfrac{p}{a} + \dfrac{q}{b} = 1$.Note: The coordinates and their order has to be taken very carefully in the formula for calculating the area of Triangle.In the formula for Area of $\Delta = \dfrac{1}{2}\left| {{x_1}({y_2} - {y_3}) + {x_2}({y_3} - {y_1}) + {x_3}({y_1} - {y_2})} \right|$.If instead of ${x_1}$ in the first term in the bracket, we take ${x_2}$ for the x coordinates and similarly if we change the order of the y-ordinates anywhere in the formula then the formula will become invalid.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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