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Home / Doubts and Solutions/KERALA/Class 11 Science/Mathematics

The vertices of triangle ABC are A(2,1),B(-3,5) and C(4,5)
Find the equation of the median through vertex B. 

Asked by Deebu 5th February 2015, 1:43 AM
Answered by Expert
Answer:
C o n s i d e r space t h e space g i v e n space v e r t i c e s space o f space t h e space t r i a n g l e comma space A open parentheses 2 comma 1 close parentheses comma space B open parentheses negative 3 comma 5 close parentheses space a n d space C open parentheses 4 comma 5 close parentheses. T h e space s i d e space o p p o s i t e space t o space t h e space v e r t e x space B space i s space C A. M e d i a n space i s space t h e space l i n e space j o i n i n g space t h e space v e r t e x space a n d space t h e space m i d p o i n t space o f space t h e space o p p o s i t e space s i d e. T h u s space m i d p o i n t space o f space C A equals open parentheses fraction numerator 2 plus 4 over denominator 2 end fraction comma fraction numerator 1 plus 5 over denominator 2 end fraction close parentheses equals open parentheses 3 comma 3 close parentheses L e t space E space b e space t h e space m i d p o i n t space o f space C A T h u s space t h e space c o o r d i n a t e s space o f space E equals open parentheses 3 comma 3 close parentheses N o w space w e space n e e d space t o space f i n d space t h e space e q u a t i o n space o f space t h e space l i n e space j o i n i n g space t h e space p o i n t s space B open parentheses negative 3 comma 5 close parentheses space a n d space E open parentheses 3 comma 3 close parentheses. E q u a t i o n space o f space t h e space l i n e space j o i n i n g space t h e space t w o space p o i n t s space i s fraction numerator x minus x subscript 1 over denominator x subscript 2 minus x subscript 1 end fraction equals fraction numerator y minus y subscript 1 over denominator y subscript 2 minus y subscript 1 end fraction T h e r e f o r e comma space e q u a t i o n space o f space B E space i s comma fraction numerator x minus open parentheses negative 3 close parentheses over denominator 3 minus open parentheses negative 3 close parentheses end fraction equals fraction numerator y minus 5 over denominator 3 minus 5 end fraction rightwards double arrow fraction numerator x plus 3 over denominator 6 end fraction equals fraction numerator y minus 5 over denominator negative 2 end fraction rightwards double arrow fraction numerator x plus 3 over denominator 3 end fraction equals fraction numerator y minus 5 over denominator negative 1 end fraction rightwards double arrow open parentheses negative 1 close parentheses open parentheses x plus 3 close parentheses equals 3 open parentheses y minus 5 close parentheses rightwards double arrow negative x minus 3 equals 3 y minus 15 rightwards double arrow x plus 3 y minus 15 plus 3 equals 0 rightwards double arrow x plus 3 y minus 12 equals 0
Answered by Expert 5th February 2015, 9:16 AM
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