First slide

Cartesian plane

Question

$Area of the triangle formed by the line x + y = 3 and the angle bisectors of the pairs of straight lines x^{2} - y^{2} + 2 y = 1 i s$

Moderate

A

2 sq. units
B

4 sq. units
C

6 sq. units
D

8 sq. units

Solution

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

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