The equation of the medians of a triangle formed by the lines x+y−6=0,x−3y−2=0 and 5x−3y+2=0, is
A
x=2,x+9y−14=0 and 7x−9y−2=0
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B
x=2,x−9y−14=0 and 7x−9y+2=0
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C
x=2,x+9y+14=0 and x−9y−2=0
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D
x=2,x−9y−14=0 and 7x+9y+2=0
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Solution
The correct option is Ax=2,x+9y−14=0 and 7x−9y−2=0 Let the given equation form a triangle ΔABC whose sides equations are: AB:x+y−6=0...(i) BC:x−3y−2=0...(ii) CA:5x−3y+2=0...(iii) So intersection point of AB,BC is B(5,1) similarly for BC,CA and CA,AB we get C(−1,−1) and A(2,4) respectively Let G be the centriod, then all the medians will pass through G only and G≡(2,43) So equation of the median passing through A will be x=2 equation of the median passing through B will be y−43=4/3−12−5(x−2)⇒x+9y=14 equation of the median passing through C will be y+1=4/3+12+1(x+1)⇒7x−9y=2