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(B) Vertex $A$ is the point of intersection of the lines $x+y=1$ and $2x+3y=6$. Solving these,we get $x = -3$ and $y = 4$. So,$A(-3, 4)$ and $O\left(\

Vedclass · Accepted Answer

$x \cos \alpha+y \sin \alpha=\frac{13}{\sqrt{17}} ; \alpha=	an ^{-1}(4)$

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(B) Vertex $A$ is the point of intersection of the lines $x+y=1$ and $2x+3y=6$. Solving these,we get $x = -3$ and $y = 4$. So,$A(-3, 4)$ and $O\left(\frac{3}{7}, \frac{22}{7}ight)$. The equation of the line $OA$ passing through $A(-3, 4)$ and $O\left(\frac{3}{7}, \frac{22}{7}ight)$ is: $y - 4 = \frac{\frac{22}{7} - 4}{\frac{3}{7} - (-3)}(x + 3)$ $y - 4 = \frac{-6/7}{24/7}(x + 3)$ $y - 4 = -\frac{1}{4}(x + 3)$ $4y - 16 = -x - 3 \Rightarrow x + 4y = 13$. The normal form of the line $x \cos \alpha + y \sin \alpha = p$ is obtained by dividing by $\sqrt{1^2 + 4^2} = \sqrt{17}$. $\frac{1}{\sqrt{17}}x + \frac{4}{\sqrt{17}}y = \frac{13}{\sqrt{17}}$. Here,$\cos \alpha = \frac{1}{\sqrt{17}}$ and $\sin \alpha = \frac{4}{\sqrt{17}}$,so $	an \alpha = 4$,which means $\alpha = 	an^{-1}(4)$. Thus,the equation is $x \cos \alpha + y \sin \alpha = \frac{13}{\sqrt{17}}$ with $\alpha = 	an^{-1}(4)$.

The vertex $A$ of a triangle lies on the lines $x+y=1$ and $2x+3y=6$. If the orthocentre of the triangle is $O\left(\frac{3}{7}, \frac{22}{7}\right)$,then the equation of $OA$ in the normal form is

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