Question
The area of the triangle formed by two rays whose combined equation is $$y = \left| x \right|$$ and the line $$x + 2y = 2$$ is :
A.
$$\frac{8}{3}{\text{ uni}}{{\text{t}}^2}$$
B.
$$\frac{4}{3}{\text{ uni}}{{\text{t}}^2}$$
C.
$$4{\text{ uni}}{{\text{t}}^2}$$
D.
$$\frac{{16}}{3}{\text{ uni}}{{\text{t}}^2}$$
Answer :
$$\frac{4}{3}{\text{ uni}}{{\text{t}}^2}$$
Solution :
The lines are $$y=x,\, y=-x$$ and $$x+2y=2$$ as shown in the figure.
Solving $$y=x,\,x+2y=2,$$ the point $$A = \left( {\frac{2}{3},\frac{2}{3}} \right)$$
$$\eqalign{ & \therefore \,\,OA = \sqrt {\frac{4}{9} + \frac{4}{9}} = \frac{{2\sqrt 2 }}{3} \cr & {\text{Solving }}y = - x,\,x + 2y = 2,{\text{ the point}} \cr & B = \left( { - 2,\,2} \right) \cr & \therefore \,\,OB = \sqrt {4 + 4} = 2\sqrt 2 \cr & \therefore \,\,{\text{ar}}\left( {\Delta OAB} \right) = \frac{1}{2}.OA.OB = \frac{1}{2}.\frac{{2\sqrt 2 }}{3}.2\sqrt 2 = \frac{4}{3} \cr} $$
The lines are $$y=x,\, y=-x$$ and $$x+2y=2$$ as shown in the figure.
Solving $$y=x,\,x+2y=2,$$ the point $$A = \left( {\frac{2}{3},\frac{2}{3}} \right)$$
$$\eqalign{ & \therefore \,\,OA = \sqrt {\frac{4}{9} + \frac{4}{9}} = \frac{{2\sqrt 2 }}{3} \cr & {\text{Solving }}y = - x,\,x + 2y = 2,{\text{ the point}} \cr & B = \left( { - 2,\,2} \right) \cr & \therefore \,\,OB = \sqrt {4 + 4} = 2\sqrt 2 \cr & \therefore \,\,{\text{ar}}\left( {\Delta OAB} \right) = \frac{1}{2}.OA.OB = \frac{1}{2}.\frac{{2\sqrt 2 }}{3}.2\sqrt 2 = \frac{4}{3} \cr} $$