The orthocentre of the triangle formed by the lines \(x+y=1,\:2x+3y=6\) and \(4x-y+4=0\) lies in

(A) first quadrant

(B) second quadrant

(C) third quadrant

(D) fourth quadrant

(A) first quadrant

(B) second quadrant

(C) third quadrant

(D) fourth quadrant

Answer: A

Let \(H(h,k)\) be orthocentre.

Let \(H(h,k)\) be orthocentre.

\(\Rightarrow(\)slopes of \(AH).(\)slope of \(BC)\)

\(\displaystyle\Rightarrow \left( \dfrac { k-\dfrac { 16 }{ 7 } }{ h+\dfrac { 3 }{ 7 } } \right) .\left( -1 \right) =-1\)

\(\displaystyle\Rightarrow k-\frac { 16 }{ 7 } =h+\frac { 3 }{ 7 } \)

\(\displaystyle\Rightarrow h-k=-\frac { 19 }{ 7 } \) ...(i)

Also, \((\)slope of \(CH).(\)slopes of \(AB)=-1\)

\(\displaystyle\Rightarrow \frac { k-4 }{ h+3 } .\left( 4 \right) =-1\Rightarrow 4k-16=-h-3\)

\(\displaystyle\Rightarrow h+4k=13\) ...(ii)

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