question_answer

Area of the triangle formed by the line \[x+y=3\]and the angle bisectors of the pairs of straight lines \[{{x}^{2}}-{{y}^{2}}+2y=1\] is

A) 2 sq. units

B) 4 sq. units

C) 6 sq. units

D) 8 sq. units

Correct Answer: A

Solution :

[a] \[{{x}^{2}}-{{y}^{2}}+2y=1\] or \[x=\pm (y-1)\] The bisectors of the above lines are x = 0 and y = 1. So, the area between \[x=0,\,\,y=1\] and \[x+y=3\] is the shaded region shown in the figure. The area is given by \[(1/2)\times 2\times 2=2\]sq. units.