The vertex $A$ of a triangle lies on the lines $x+y=1$ and $2 x+3 y=6$. If the orthocentre of the triangle…
The vertex $A$ of a triangle lies on the lines $x+y=1$ and $2 x+3 y=6$. If the orthocentre of the triangle is $O\left(\frac{3}{7}, \frac{22}{7}\right)$, then the equation of $O A$ in the normal form is
Vertex $A$ is point of intersection of the lines $x+y=1$ and $2 x+3 y=6$
Point of intersection $=(-3,4)$
So, $A(-3,4)$ and $O\left(\frac{3}{7}, \frac{22}{7}\right)$
Equation of line passing through two points
$
\begin{aligned}
y-4 & =\frac{\left(\frac{22}{7}-4\right)}{\frac{3}{7}+3}(x+3) \\
y-4 & =-\frac{6}{24}(x+3) \\
x+4 y & =13
\end{aligned}
$
Normal form of line $x \cos \alpha+y \sin \alpha=P$
Where $P=\frac{13}{\sqrt{17}}, \cos \alpha=\frac{1}{\sqrt{17}}, \sin \alpha=\frac{4}{\sqrt{17}}$
$
\Rightarrow \quad \tan \alpha=\frac{4}{1}
$
Hence, required equation is
$
x \cos \alpha+y \sin \alpha=\frac{13}{\sqrt{17}} ; \alpha=\tan ^{-1}(4) \text {. }
$