find the area of the triangle formed by the line x + y = 3 and angle bisectors of the pair of straight lines x squared minus y squared plus 2 y equals 1.

Asked by Topperlearning User | 12th Aug, 2016, 11:48: PM

Expert Answer:

H e r e space x squared minus y squared plus 2 y minus 1 equals 0 rightwards double arrow x squared minus left parenthesis y minus 1 right parenthesis squared equals 0 rightwards double arrow left parenthesis x minus y plus 1 right parenthesis left parenthesis x plus y minus 1 right parenthesis equals 0
x minus y plus 1 equals 0 space a n d space x plus y minus 1 equals 0 space a r e space t h e space t w o space s t r a i g h t space l i n e s.
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N o w space t h e space e q u a t i o n space o f space a n g l e space b i s e c t o r s space o f space t h e s e space t w o space s t r a i g h t space l i n e s space w i l l space b e space g i v e n space b y
fraction numerator x minus y plus 1 over denominator square root of 2 end fraction equals plus-or-minus fraction numerator x plus y minus 1 over denominator square root of 2 end fraction
w i t h space t h e space plus space s i g n space w e space w i l l space g e t space y equals 1
w i t h space t h e space minus space s i g n space w e space w i l l space g e t space x equals 0
s o space b i s e c t o r s space a r e space x equals 0 space a n d space y equals 1
So, area between x + y = 3, x = 0 and y = 1 is given by
A =

Answered by  | 13th Aug, 2016, 01:48: AM