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

Option: 1

2 sq. units


Option: 2

4 sq. units


Option: 3

6 sq. units


Option: 4

8 sq. units


Answers (1)

best_answer

Here,
\mathrm{x^{2}-y^{2}+2 y=1}
\mathrm{\Rightarrow \mathrm{x}^{2}-\left(\mathrm{y}^{2}-2 \mathrm{y}+1\right)=0}
\mathrm{\Rightarrow \mathrm{x}^{2}=(\mathrm{y}-1)^{2}}

\mathrm{or \, \mathrm{x}=\mathrm{y}-1\; and \; \mathrm{x}=-\mathrm{y}+1}

Which could be graphically as shown in figure.

 

Which gives angle bisector as
\mathrm{\mathrm{y}=1\: and \: \mathrm{x}=0}

\therefore  area of region bounded by
\mathrm{x}+\mathrm{y}=3, \mathrm{x}=0$ and $\mathrm{y}=1

=\frac{1}{2} \times 2 \times 2=2 \text { sq. units }
 


 

Posted by

Anam Khan

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