The point 'B' is the image of 'A' in the line x+y+4=0x+y+4=0 and 'C' is the image of 'B' in the line 2 x-y+7=0,2x−y+7=0, if A=(1,6)A=(1,6) then the circumcentre of the triangle \mathrm{ABC}ABC is
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Step-by-step explanation:
ANSWER
5x+3y−2=0
(3x−y−4=0)×3
14x−10=0
x=
7
5
5×
7
5
+3y−2=0
3y+
7
−14+15
=0
3y=
7
−1
⇒y=−1/21
P(
7
5
,
21
−1
)
Similarity x−y+1=0
−
2
x−
+
y
−
+
2
=0
−x+1=0,x=1 and y=2
Q(I,2)
equation of line passing through P & Q
5x+3y02=0
3x−y−4=0)×3
14x−14=0
x=1
y=−1
P(I,−)
2X−Y−2=0
−
x
−
+
y
−
1
=0
x−3=0
x=3
y=4
Q(3,4)
Equation of line PQ
y+1=
3−1
4+1
(x−1)
y+1=
2
5
(x−1)
2y+2=5x−5
5x−2y−7=0
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