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Byju's Answer
Standard XII
Mathematics
Vertices of Hyperbola
Show that the...
Question
Show that the following points are coplanar.
(i) (0, −1, 0), (2, 1, −1), (1, 1, 1) and (3, 3, 0)
(ii) (0, 4, 3), (−1, −5, −3), (−2, −2, 1) and (1, 1, −1)
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Solution
(i) The equation of the plane passing through points (0, −1, 0), (2, 1, −1), (1, 1, 1) is given by
x
-
0
y
+
1
z
-
0
2
-
0
1
+
1
-
1
-
0
1
-
0
1
+
1
1
-
0
=
0
⇒
x
y
+
1
z
2
2
-
1
1
2
1
=
0
⇒
4
x
-
3
y
+
1
+
2
z
=
0
⇒
4
x
-
3
y
+
2
z
-
3
=
0
Substituting the last point (3, 3, 0) (it means x = 3; y = 3; z = 0) in this plane equation, we get
4
3
-
3
3
+
2
0
-
3
=
0
⇒
12
-
12
=
0
⇒
0
=
0
So, the plane equation is satisfied by the point (3, 3, 0).
So, the given points are coplanar.
(ii) The equation of the plane passing through (0, 4, 3), (−1, −5, −3), (−2, −2, 1) is
x
-
0
y
-
4
z
-
3
-
1
-
0
-
5
-
4
-
3
-
3
-
2
-
0
-
2
-
4
1
-
3
=
0
⇒
x
y
-
4
z
-
3
-
1
-
9
-
6
-
2
-
6
-
2
=
0
⇒
-
18
x
+
10
y
-
4
-
12
z
-
3
=
0
⇒
9
x
-
5
y
-
4
+
6
z
-
3
=
0
⇒
9
x
-
5
y
+
6
z
+
2
=
0
Substituting the last point (1, 1, -1) (it means x = 1; y = 1; z=-1) in this plane equation, we get
9
1
-
5
1
+
6
-
1
+
2
=
0
⇒
4
-
4
=
0
⇒
0
=
0
So, the plane equation is satisfied by the point (1, 1, -1).
So, the given points are coplanar.
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If the points (2, 1) and (1, −2) are equidistant from the point (x, y), show that x + 3y = 0.