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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
I quadrant
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B
II quadrant
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C
III quadrant
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D
IV quadrant
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Solution

The correct option is C II quadrant

We have,

x+y=1......(1)

2x+3y=6......(2)

4xy=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

4xy=4

4x+6y=12

4xy=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=6487

2x=42487

2x=67

x=37

So, (x2,y2)=(37,167)

From equation (3) and (1) to, and we get,

4xy=4

x+y=1

On adding and we get,

5x=3

x=35

Put the value of x in equation (4) and we get,

4xy=4

4×35y=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)=(1051521105,140+80+56105)

(x,y)=(141105,276105)

(x,y)=(,+)IIndquadrent

Hence, this is the answer.

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