Two vertices of a Triangle are (1,3) and (4,7). If its orthocentre lies on the line x+y=3, then the locus of third vertex is

Let A be (x,y) and orthocentre be $\left(\alpha ,3-\alpha \right)$

$\because BO\perp BC\Rightarrow \frac{y-7}{x-4}.\frac{\alpha }{1-\alpha }=-\mathrm{1.........}\left(1\right)$

$\because AO\perp BC\Rightarrow \frac{y-3+\alpha }{x-\alpha }\times \frac{4}{3}=-\mathrm{1.........}\left(2\right)$

Eliminating $\alpha$ from equation (1) and (2) we get option (3)