Question

Sides of a triangle are represented by the line x+y=3 and angle bisectors of the pair of straight lines x^2-y^2+2y=1 then area of triangle is

Answer 1

First, observe that just like @Nicholas said, the equation x2−y2+2y=1x2−y2+2y=1defines two lines:

x2−y2+2y=1⟺x2=(y−1)2⟹{l1:l2:y=x+1y=−x+1x2−y2+2y=1⟺x2=(y−1)2⟹{l1:y=x+1l2:y=−x+1

The slope of the first one is π4=45º,π4=45º,and for the second one the slope is 3π4=135º,3π4=135º, i.e. they are perpendicular.

Equations of bisecting lines can be found by adding and subtracting equations of original lines:

{b1:b2:x=0y=1{b1:x=0b2:y=1

Points of intersection of these lines with each other and with the line l0: y=3−xl0: y=3−x are

ABC:=l0∩b1:=b1∩b2:=l0∩b1⟺⟺⟺{y=3−xx=0{x=0y=1{y=3−xy=1⟹⟹⟹ABC=(0,3)=(0,1)=(2,1)A:=l0∩b1⟺{y=3−xx=0⟹A=(0,3)B:=b1∩b2⟺{x=0y=1⟹B=(0,1)C:=l0∩b1⟺{y=3−xy=1⟹C=(2,1)

Observe that b1⊥b2b1⊥b2 so that △ABC△ABCis right triangle. Therefore the area SS of △ABC△ABC is just a half of product of legsABAB and BCBC:

S△ABC=12∥∥(0,1)−(0,3)∥∥∥∥(2,1)−(0,1)∥∥=2S△ABC=12‖(0,1)−(0,3)‖‖(2,1)−(0,1)‖=2