Question

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

• A. $-4$
• B. $0$
• C. $5$
• D. $6$

Answer 1

The correct option is A $-4$

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

Then, area of triangle $=\frac{1}{2}|x_1(y_2.y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$

Given area $=2$

$\Rightarrow |x_2(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|=4$

Substituting values

$|x(1-2)+0(2-0)+2(0-1)|=4$

$|-2-x|=4$

$\Rightarrow 2+x=\pm 4;x=-6,+2$

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

(Option $A$)