Question

b) Find the equation of line which
passes through the point of
intersection of the lines x+2y-3=0
and 3x+4y-5=0 and which is
perpendicular to the line x-3y+5=0.​

Answer 1

Answer:

3x+4y−5=0 is the required equation of line.

Answer 2

Answer:

First , we will find out the point of intersection of the lines x+2y-3=0 and 3x+4y-5=0.

Let point p(a,b) be point of intersection.

➝ x+2y-3=0

➝ a+2b-3=0

➝ a = 3-2b...... (1)

➝ 3x+4y-5=0

➝ 3a+4b-5=0

➝ a = (5-4b)/3.....(2)

From (1) and (2) :-

➝ 3-2b = (5-4b)/3

➝( 3-2b)3 = (5-4b)

➝ 9-6b = 5-4b

➝ 9-5 = 6b-4b

➝ 4 = 2b

➝ 2 = b

Now we will find the value of a.

=> a = 3-2b

=> a = 3-(2×2)

=> a = 3-4

=> a = -1

Point of intersection of the lines x+2y-3=0 and 3x+4y-5=0 = (-1,2)

Now , we will find Slope of line x-3y+5=0.

➝ x-3y+5=0

➝ -3y= -x -5

➝ y= (x + 5)/3

➝ y= (x/3) + ( 5/3)

Now we have transformed equation of line x-3y+5=0 as y= (x/3) + ( 5/3) which is of the form y = mx + c.

Therefore, Slope of line x-3y+5=0 = 1/3

Let. slope of line which passes through the point of intersection of the lines x+2y-3=0

and 3x+4y-5=0 and which is perpendicular to the line x-3y+5=0 be n.

➝ n × (1/3) = -1

[ product of slope of two perpendicular lines = -1 ]

➝ n = -1 × 3

➝ n = - 3

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Now , we have to find equation of line which passes through point (-1,2) and has slope -3.

➝ y - 2 = (x+1)(-3)

➝ y - 2 = -3x-3

➝ y + 3x= -3+2

➝ y + 3x= - 1

➝ y + 3x + 1=0

Answer :

y + 3x + 1=0

Additional Information :

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x

2

−x

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y

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−y

1