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Question : 88 of 120
Marks:
+1 ,
-0
(1, 2, 3) (–3, 2, 0) (3, –2, 0) (3, 2, 0) (3, 2, 1)
Solution:
Let the points are
A ( 2 , 0 , 3 ) , B ( 0 , 3 , 2 ) and
D ( 0 , 0 , 1 ) We know that
Z -coordinate of every point an
x y -plane is zero so let
p ( x , y , 0 ) be a point on
x y -plane such that
P A = P B = P C .
Now,
P A = P B ⇒ P A 2 = P B 2 ⇒ ( x − 2 ) 2 + ( y − 0 ) 2 + ( 0 − 3 ) 2 = ( x − 0 ) 2 + ( y − 3 ) 2 + ( 0 − 2 )
⇒ ‌ ‌ 4 x − 6 y = 0 ⇒ 2 x − 3 y = 0 ....(i)
and,
‌ ‌ P B = P C ⇒ ‌ ‌ P B 2 = P C 2 ⇒ ( x − 0 ) 2 + ( y − 3 ) 2 + ( 0 − 2 ) 2 = ( x − 0 ) 2 + ( y − 0 ) 2 + ( 0 − 1 ) 2
⇒ ‌ ‌ − 6 y + 12 = 0 ⇒ ‌ ‌ y = 2 ....(ii)
Putting
y = 2 in Eq. (i), we get
x = 3 Hence, the required point is
( 3 , 2 , 0 ) © examsiri.com
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