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# Question and Answer

## QuestionMathsClass 11

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

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)