(b) Find the point equidistant from the points (1, 2), (3, 4) and (5, -6). (c) Find the cicumcentre of the triangle whose vertices are (4,0), (0, 4) and (-272, -2,2). (d) Find the centre of the circle passing through the points (3,0), (2, 5) and (-2,2, -1).
Answers
Answer:
Solution
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Correct option is D)
∵ Nine point centre is the midpoint of orthocentre and circumcentre.
∵A≡(1,2),B≡(2,3) and C≡(4,3)
∴ Centroid G≡(
3
1+2+4
,
2
2+3+3
)=(
3
7
,
3
8
)
Let O
′
(α,β) be the orthocentre
As O
′
D⊥ to BC we have
∴ Slope of O
′
A×Slope of BC=−1
⇒
α−1
β−2
×
2
0
=−1 or α−1=0
∴α=1
As O
′
E⊥ to AB we have
Also, slope of O
′
C×slope of AB=−1
⇒
α−4
β−3
×
2−1
3−2
=−1
Substituting for α=1 from above we get
⇒β−3=−(α−4)=−1(1−4)=3
∴β=3+3=6
∴ Orthocentre O
′
≡(1,6)
∵ Centroid G divides orthocentre O
′
and circumcentre C
′
in the ratio 2:1 internally.
If C
′
≡(x
1
,y
1
) then O
′
(1,6):G(
3
7
,
3
8
)=2:1
G(
3
7
,
3
8
):C
′
(x
1
,y
1
)=1:1
∴G divides the line O
′
C
′
in the ratio 2:1 which is given by the section formula: (
m+n
mx
2
+nx
1
,
m+n
my
2
+ny
1
)
∴
3
7
=
2+1
2.x
1
+1.1
⇒x
1
=3 and
3
8
=
2+1
2.y
1
+1.6
On simplification, we get 2y
1
+6=8 or y
1
=1
∴C
′
≡(3,1)
Then , O
′
C
′
=∣1−3∣+∣6−1∣=2+5=7units
Hence,Nine point centre=(
2
1+3
,
2
6+1
)=(2,
2
7
)