The distance from the origin to the orthocent | vedclass

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(D) The given lines are $x+y-1=0$ and $6x^2-13xy+5y^2=0$.Factorizing the second equation:$6x^2-10xy-3xy+5y^2=0$$(2x-y)(3x-5y)=0$So,the lines are $2x-y

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$\frac{11\sqrt{2}}{24}$

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(D) The given lines are $x+y-1=0$ and $6x^2-13xy+5y^2=0$.Factorizing the second equation:$6x^2-10xy-3xy+5y^2=0$$(2x-y)(3x-5y)=0$So,the lines are $2x-y=0$ and $3x-5y=0$.Let the vertices of the triangle be $O(0,0)$,$A$,and $B$.Solving $2x-y=0$ and $x+y-1=0$,we get $A = (1/3, 2/3)$.Solving $3x-5y=0$ and $x+y-1=0$,we get $B = (5/8, 3/8)$.Let the orthocentre be $H(h, k)$.The altitude from $B$ to $OA$ is perpendicular to $y=2x$. Its equation is $y - 3/8 = -1/2(x - 5/8) \Rightarrow 4x + 8y - 6 = 0$.The altitude from $A$ to $OB$ is perpendicular to $y=3/5x$. Its equation is $y - 2/3 = -5/3(x - 1/3) \Rightarrow 15x + 9y - 11 = 0$.Solving these two altitude equations:$4h + 8k = 6$ and $15h + 9k = 11$.Multiplying the first by $9$ and second by $8$:$36h + 72k = 54$$120h + 72k = 88$Subtracting gives $84h = 34 \Rightarrow h = 17/42$.Substituting $h$ in $4h + 8k = 6$: $4(17/42) + 8k = 6$ $\Rightarrow 34/42 + 8k = 6$ $\Rightarrow 8k = 6 - 17/21 = 109/21$ $\Rightarrow k = 109/168$.The distance from origin to $(h, k)$ is $\sqrt{h^2+k^2} = \sqrt{(17/42)^2 + (109/168)^2} = \frac{11\sqrt{2}}{24}$.

The distance from the origin to the orthocentre of the triangle formed by the lines $x+y-1=0$ and $6x^2-13xy+5y^2=0$ is

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