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Question : 17 of 75
Marks:
+1,
-0
Solution:
Equation of line in the internal bisector of
and
is
(√3+1)+(√3+1)⇒ line will be
y=x⇒x−y=0‌D=|‌| a−(1−a) |
| √a2+(1−a)2 |
|=‌‌(2a−1)2=‌(a2+(1−a)2)‌⇒2(4a2−4a+1)=81a2+81a2−162a−81‌⇒162a2−162a+81−8a2+8a−2=0‌⇒154a2−154a+79=0‌‌ Sum of values ‌=−‌=1
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