Find the equation of the locus of a point which is equidistant from the point (1, 3)
and X-axis.
Answers
Answered by
1
Answer:
Answer
Let A(h,k) be the point equidistant from B(1,3) and y=0
=>
(h−1)
2
+(k−3)
2
=
1
k
=>k
2
−2h+1+k
2
−6k+9=k
2
=>h
2
−2h−6k+10=0
locus of point A=>x
2
−2x−6y+10=0
Answered by
2
Answer:
0
Step-by-step explanation:
Let A(h,k) be the point equidistant from B(1,3) and y=0
=>
(h−1)
2
+(k−3)
2
=
1
k
=>k
2
−2h+1+k
2
−6k+9=k
2
=>h
2
−2h−6k+10=0
locus of point A=>x
2
−2x−6y+10=0
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