Find the equations of the straight lines passing through (1, 1) and which are at a distance of 3 units
from (-2,3).
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Answer:
ANSWER
Suppose that the equation of line is
x+by+c=0.......(1)
The line passes through points (1,1)
Then,
1+b+c=0
⇒c=−(b+1)
The equation of the line is x+by−(b−1)=0.
The line is at a distance of 3 units from
(−2,3).
Now,
∣
∣
∣
∣
∣
1+b
2
−2+3b−(b+1)
∣
∣
∣
∣
∣
=3
⇒
∣
∣
∣
∣
∣
1+b
2
=3
2b−3
∣
∣
∣
∣
∣
On squaring both side and we get,
⇒
1+b
2
4b
2
−12b+9
=9
⇒4b
2
−12b+9=9+9b
2
⇒5b
2
+12b=0
⇒b(5b+12)=0
⇒b=0,5b+12=0
⇒b=0,b=−
5
12
Hence, the equation of ,line is x+by−(b+1)=0
If b=0 then,
Equation of line is x−1=0
If b=−
5
12
Then equation of line is
x−(
5
12
)+
5
7
=0
5x−12y+7=0
Hence, this is the answer.
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