Distance between A and B $= \sqrt {{{(2)}^2} + {{( - 1)}^2}} = \sqrt 5$ Area of triangle = $5\sqrt 5$ $\Rightarrow {1 \over 2} \times \sqrt 5 \times x = 5\sqrt 5$ $\Rightarrow x = 10$ Slope of AB, ${m_{AB}} = {{1 - 2} \over {3 - 1}} = - {1 \over 2}$ AC is perpendicular to AB. $\therefore$ ${m_{AB}}\,.\,{m_{AC}} = - 1$ $\Rightarrow {m_{AC}} = 2 = \tan \theta$ So, $\sin \theta = {2 \over {\sqrt 5 }}$, $\cos \theta = {1 \over {\sqrt 5 }}$ $\therefore$ $a = {x_A} + r\cos \theta$ $= 1 + 10 \times {1 \over {\sqrt 5 }}$ $= 1 + 2\sqrt 5$