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Question

# The centroid of the triangle formed by the lines x+y=1, 2x+3y=6 and 4x-y+4=0 lines in

A
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Solution

## The correct option is C II quadrantWe have,x+y=1......(1) 2x+3y=6......(2) 4x−y=−4......(3)From equation (1) and (2) to and we get, (x+y=1)×2 2x+3y=6 2x+2y=2 2x+3y=6 On subtracting and we get, y=4 put in (1) and we get, 2x+2y=2 2x+2(4)=2 2x+8=2 2x=−6 x=−3 Then, (x1,y1)=(−3,4) From equation (2) and (3) to, and we get (2x+3y=6)×2 4x−y=−4 4x+6y=12 4x−y=−4 On subtracting and we get, 7y=16 y=167 Put the value of y in equation (2) and we get, 2x+3y=6 2x+3×167=6 2x+487=6 2x=6−487 2x=42−487 2x=−67 x=−37 So, (x2,y2)=(−37,167) From equation (3) and (1) to, and we get, 4x−y=−4 x+y=1 On adding and we get, 5x=−3 x=−35 Put the value of x in equation (4) and we get, 4x−y=−4 4×−35−y=−4 −125+4=y y=−12+205 y=85 So, (x3,y3)=(−35,85) Then the orthocenter is (x,y)=(x1+x2+x33,y1+y2+y33) (x,y)=(−3+(−37)+(−35))3,4+167+853 (x,y)=(−105−15−21105,140+80+56105) (x,y)=(−141105,276105) (x,y)=(−,+)IIndquadrent Hence, this is the answer.

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