Question

9. 1) Find the equation of the straight line parallel to 4x - 3y+15=0 and passing through
the intersection of the lines 2x – 3y = 5 and 3x + 2y = 1.
ii) Find the ratio in which the line 3x - 2y +5 = 0 divides the line joining the points
(2.3) and (3.-5).​

Answer 1

Answer:

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).

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ANSWER

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