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

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Q. A triangle has two of its vertices at $\left(0 , 1\right)$ and $\left(2 , 2\right)$ in the Cartesian plane. Its third vertex lies on the $x$ -axis. If the area of the triangle is $2$ square units, then the sum of the square of the possible values of abscissa of the third vertex is

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Solution:

Solution
Let, the third vertex be $\left(a , 0\right)$
Area of triangle, $\Delta =\frac{1}{2}\begin{vmatrix} 0 & 1 & 1 \\ 2 & 2 & 1 \\ a & 0 & 1 \end{vmatrix}=2$
$\therefore \begin{vmatrix} 0 & 0 & 1 \\ 2 & 1 & 1 \\ a & -1 & 1 \end{vmatrix}=4$
$\Rightarrow \left|a + 2\right|=4$
i.e. $a=2$ or $a=-6$
Hence, the sum of the square of the possible values of abscissa $=40$