7. Find the angle P of the triangle whose vertices are P(0,-1,-2), Q(3, 1, 4) and R(5, 7, 1).
Answers
Answer:
We know that the distance between the two points (x
1
,y
1
) and (x
2
,y
2
) is
d=
(x
2
−x
1
)
2
+(y
2
−y
1
)
2
Let the given vertices be A=(8,−4), B=(9,5) and C=(0,4)
We first find the distance between A=(8,−4) and B=(9,5) as follows:
AB=
(x
2
−x
1
)
2
+(y
2
−y
1
)
2
=
(10−2)
2
+(1−1)
2
=
8
2
+0
2
=
64
=8
Similarly, the distance between B=(9,5) and C=(0,4) is:
BC=
(x
2
−x
1
)
2
+(y
2
−y
1
)
2
=
(6−10)
2
+(9−1)
2
=
(−4)
2
+8
2
=
16+64
=
80
=
4
2
×5
=4
5
Now, the distance between C=(0,4) and A=(8,−4) is:
CA=
(x
2
−x
1
)
2
+(y
2
−y
1
)
2
=
(6−2)
2
+(9−1)
2
=
4
2
+8
2
=
16+64
=
80
=
4
2
×5
=4
5
We also know that If any two sides have equal side lengths, then the triangle is isosceles.
Here, since the lengths of the two sides are equal that is BC=CA=4
5
Hence, the given vertices are the vertices of an isosceles triangle.