Question

The straight lines 3x + 4y = 5 and 4x - 3y = 15 -intersect at the point A. On these lines, the points B and C are chosen so that AB = AC. Find possible equation of the line BC passing through the point (1, 2).

Answer 1

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 $\pm {45}^{\circ }$ 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)
$\therefore$ y - 2 = m ( x - 1)
where m is slope of any of bisectors given by
$\frac{3x+4y-5}{5}=\pm \frac{4x-3y-15}{5}$
or x - 7y + 13 = 0 or 7x + y - 20 = 0
$\therefore$ m = 1 / 7 or - 7
putting in (1) , the required lines are 7x + y - 9 = 0
and x - 7y + 13 = 0 as found above