Math, asked by devapradeep114, 5 months ago

find the equation of the straight line joining the point of intersection of 3x+y+2=0 and x-2y-4= to the point of intersection of 7x-3y=-12 and 2y=x+3​

Answers

Answered by Anonymous
1

Answer:Step 1.

( Finding the point of intersection of the first pair of lines )

The first pair of straight lines are

3x + y + 2 = 0 .....(1)

x - 2y - 4 = 0 .....(2)

Multiplying (1) no. equation by 2 and adding with (2) no. equation, we get

6x + 2y + 4 + x - 2y - 4 = 0

⇒ 7x = 0

⇒ x = 0

Putting x = 0 in (1) no. equation, we get

0 + y + 2 = 0

⇒ y = - 2

∴ (0, - 2) is the point of intersection of the first pair of straight lines.

Step 2.

( Finding the point of intersection of the second pair of lines )

The second pair of straight lines are

7x - 3y = - 12 .....(iii)

2y = x + 3 ⇒ x = 2y - 3 .....(iv)

Using (iv), from (iii), we get

7 (2y - 3) - 3y = - 12

⇒ 14y - 21 - 3y = - 12

⇒ 11y = 21 - 12 = 9

⇒ y = 9/11

Putting y = 9/11 in (iv) no. equation, we get

x = 2 (9/11) - 3

⇒ x = 18/11 - 3

⇒ x = (18 - 33)/11

⇒ x = - 15/11

∴ (- 15/11, 9/11) is the point of intersection of the second pair of straight lines.

Step 3.

( Finding the required line of intersection )

The two points of intersection are (0, - 2) and (- 15/11, 9/11).

Thus, the line passing through the above points is

(x - 0)/{0 - (- 15/11)} = {y - (- 2)}/(- 2 - 9/11)

⇒ x/(15/11) = (y + 2)/{(- 22- 9)/11}

⇒ 11x/15 = (y + 2)/(- 31/11)

⇒ 11x/15 = - 11 (y + 2)/31

⇒ 31 * 11x = - 15 * 11 (y + 2)

⇒ 341x = - 165y - 330

⇒ 341x + 165y + 330 = 0 (Ans.)

Step-by-step explanation:

Answered by Anonymous
1

Answer:

here's ur answer dude

Step-by-step explanation:

Line passing through point of intersecting of L

1

,L

2

is given by L

1

2

=0

⇒(3x+2y+4)+λ(2x+5y−1)=0

⇒x(3+2λ)+y(2+5λ)+4−λ=0

Distance of (2,−1) to this line,

(3+2λ)

2

+(2+5λ)

2

2(3+2λ)−(2+5λ)+4−λ

=2

⇒6+4λ−2−5λ=2

29λ

2

+32λ+13

⇒−2λ+8=2

29λ

2

+32λ+13

⇒λ

2

+16−8d=29λ

2

+32λ+13

⇒28λ

2

+40λ−3=0⇒28λ

2

+42λ−2λ−3=0

14λ(2λ+3)−(2λ+3)=0

(14λ−1)(2λ+3)=0⇒λ=

14

1

or λ=

2

−3

L:x(3+

7

1

)+y(2+

14

5

)+4−

14

1

=0

44x+33y+55=0⇒

4x+3y+5=0

or x(3−3)+y(2−

2

15

)+4+

2

3

=0

⇒y(−11)+11=0⇒

y=1

hope it helps

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