Math, asked by Anonymous, 6 days ago

{SOLVE IT }
Find the equation of the straight line perpendicular to the line 9x+3y=0 and passing through the point of intersection of the lines x+8y-1=0 and x-3y+4=0.​

Answers

Answered by portugazace
3

Answer:

x + 3y + 2 = 0

Step-by-step explanation:

x+6y-1=0 .. (1)

x-3y+4=0 .. (2)

y = 5/11 = 0.454

substitute y in 2 we get

x = -2.639

Point of intersection of above both lines is (0.454, -2.639)

Given straight line

9x + 3y = 0

Slope of line m1 = 3

Slope of line m2 = - 1/3

 

​So, the equation of line passes through intersection point

y + 1 = -1/3(x + 1)

3y + 3 = -x + 1

x + 3y + 2 = 0

Answered by Anonymous
4

Answer:

let the required line be y=mx+c

its perpendicular to 4x−2y=3

Slope of 4x−2y=3 ⇒ 4x−2y=3⇒ m

1

=

2

4

=2

m

1

×m=−1

2×m=−1

m=−

2

1

x+c

∴ y=−

2

1

x+c

Passes through intersection of 5x−8y+23=0 and 7x+6y−71=0

3(5x−8y+23)=0⇒ 15x−24y+69=0

64(7x+6y−71)=0 ⇒ 28x+24y−284=0

---------------------------------------------------------------------------------------

43x=284−69

43x=215

x=

43

215

x=5

5(5)−8y+23=0

8y=48⇒y=6

meeting point P=(5,6)

P passes through y=mx+c

⇒ 6=

2

1

(5)+c

c=6+

2

5

=

2

17

∴ y=−

2

x

+

2

17

⇒ 2y+x−17=0

Equation of the line is 2y+x−17=0

THANK YOU!

Similar questions