A line parallel to the straight line 3x – 4y - 2 = 0
and at a distance of 4 units from it is
(1) 3x – 4y + 20 = 0
(2) 4x – 3y + 12 = 0
(3) 3x – 4y + 18 = 0
(4) 3x – 4y - 22 = 0
Answers
Answer:he given lines are perpendicular and as AB = AC , Therefore △ ABC is art . angled isosceles . Hence the line BC through ( 1 , 2) will make an angles of ±45
∘
with the given lines . Its equations is y - 2 = m (x - 1) where m = 1 / 7 and -7 as in .Hence the possible equations are 7x + y - 9 = 0 and x - 7y + 13 = 0
Alt :
The two lines will be parallel to bisectors of angle between given lines and they pass through ( 1, 2)
∴ y - 2 = m ( x - 1)
where m is slope of any of bisectors given by
5
3x+4y−5
=±
5
4x−3y−15
or x - 7y + 13 = 0 or 7x + y - 20 = 0
∴ m = 1 / 7 or - 7
putting in (1) , the required lines are 7x + y - 9 = 0
and x - 7y + 13 = 0 as found above
Step-by-step explanation:
Step-by-step explanation:
The given lines are perpendicular and as AB = AC , Therefore △ ABC is art . angled isosceles . Hence the line BC through ( 1 , 2) will make an angles of ±45
∘
with the given lines . Its equations is y - 2 = m (x - 1) where m = 1 / 7 and -7 as in .Hence the possible equations are 7x + y - 9 = 0 and x - 7y + 13 = 0
Alt :
The two lines will be parallel to bisectors of angle between given lines and they pass through ( 1, 2)
∴ y - 2 = m ( x - 1)
where m is slope of any of bisectors given by
5
3x+4y−5
=±
5
4x−3y−15
or x - 7y + 13 = 0 or 7x + y - 20 = 0
∴ m = 1 / 7 or - 7
putting in (1) , the required lines are 7x + y - 9 = 0
and x - 7y + 13 = 0 as found above
