A triangle has two of its vertices at (0, 1) and (2, 2) in the cartesioan plane. Its third vertex lies on the x-axis. If the area of the triangle is 2 square units then the sum of the possible abscissae of the third vertex, is -

Let third vertex be $(x, 0)$ other given as $(0,1)(2,2)$

Then, area of triangle

$\begin{array}{l} =\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2} \cdot \mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right| \\ \text { Given area }=2 \\ \Rightarrow\left|\mathrm{x}_{2}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right|=4 \end{array}$

Substituting values

$\begin{array}{l} |\mathrm{x}(1-2)+0(2-0)+2(0-1)|=4 \\ |-2-\mathrm{x}|=4 \\ \Rightarrow 2+\mathrm{x}= \pm 4 ; \mathrm{x}=-6,+2 \end{array}$

sum of possible abscissae $=-6+2=-4$