Question

find the equation or lines which pass through the origin and makes an angle of 45° with the line 3x-y=6​

Answer 1

Answer:

2y=x

Step-by-step explanation:

The equation of any line is written as

Y= mX+C , where M is the slop of that line.

If the line passes through origin the constant (c) is zero

here given that a line passes through origin so the equation of this line must be written as Y=mX

another line is given as 3x-y=6

or y= 3x-6 , so here slop is 3

Now we know that to find the angle between any two line , we calculate this as TanФ= (M2-M1)/(1+M1*M2)

as here angle given is 45° , so Tan45= (M2-M1)/(1+M1*M2)

                                      since Tan45°= 1

so, 1+M1*M2=M2-M1

or put M2 value as 3

1+3*M1= 3-M1

i.e 4*M1= 2

or M1= 1/2

Thus equation of first line be written as Y= M*X

                                                                  Y= (1/2)*X

                                                                  2Y=X

Answer 2

Answer:

2y = x

y = -2x

Step-by-step explanation:

Tan (Angle between lines) = | (m₁ - m₂ / (1 + m₁m₂)|

3x-y=6​ => y = 3x - 6  => m₁ =  3

Tan 45° = | (3 - m₂)/(1 + 3m₂)|

=> 1 = |  (3 - m₂)/(1 + 3m₂)|

case 1 :

 (3 - m₂)/(1 + 3m₂) = 1

=> 3 - m₂ = 1 + 3m₂

=> 4m₂ = 2

=> m₂= 1/2

y = x/2 + c

as it passes through origin so c = 0

=> 2y = x

Case 2 :

 (3 - m₂)/(1 + 3m₂) = -1

=> 3 - m₂ = -1 - 3m₂

=> 2m₂ = -4

=> m₂= -2

y = -2x