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

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Aicharya · Accepted Answer

Given lines are $x+y-1=0$ and $6 x^2-13 x y+5 y^2=0$

$ \begin{array}{lc} \Rightarrow & 6 x^2-10 x y-3 x y+5 y^2=0 \\ \Rightarrow & 2 x(3 x-5 y)-y(3 x-5 y)=0 \\ \Rightarrow & (2 x-y)(3 x-5 y)=0 \\ \Rightarrow & 2 x-y=0 \\ \text { or } & 3 x-5 y=0 \end{array} $ Let orthocentre be $(h, k)$. Slope of $O P \times$ slope of $A B=-1$ $ \frac{k}{h} \times-1=-1 $

Now, slope of $O B \times$ slope of $A D=-1$ $ \begin{array}{ll} \Rightarrow & 2 \times\left(\frac{\frac{3}{8}-k}{\frac{5}{8}-h}\right)=-1 \Rightarrow \frac{3}{4}-2 h=h-\frac{5}{8} \\ \Rightarrow & \frac{3}{4}+\frac{5}{8}=3 h \Rightarrow \frac{6+5}{8}=3 h \\ \Rightarrow & \frac{11}{8}=3 h \Rightarrow h=\frac{11}{24} \\ \therefore & k=\frac{11}{24} \end{array} $ Now, $\quad O P=\sqrt{h^2+k^2}$ $ =\sqrt{\left(\frac{11}{24}\right)^2+\left(\frac{11}{24}\right)^2}=\frac{11}{24} \sqrt{2} $

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

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