4x+3y=15 and x-y=2 find the point of intersection
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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
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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
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