Question

57. A ray travelling along the line 3x - 4y =5
after being reflected from a line 1 travel along
the line 5x+12y=13. Then the equation of
the line / is
A) x+8y=0 B) x = 8y
C) 32x + 4y = 65 D) 32x - 4y +65=0​

Answer 1

Answer:

Given equation of incident ray is

3x−4y=5

Given equation of reflected ray is

5x+12y=13

Point of intersection of these line is (2,  

4

1

​  

)

Now the normal at this point will be the bisector to the angle between these lines .

So, the equation of normal is  

5

3x−4y−5

​  

=±  

13

5x+12y−13

​  

 

⇒39x−52y−65=±(25x+60y−65)

So, we get the equation of normals as  

x=8y and 32x+4y−65=0

So, the equation of lines are x=8y and 32x+4y−65=0

solution

Step-by-step explanation:

Answer 2

Answer:

Step-by-step explanation:

Given equation of incident ray is

3x−4y=5

Given equation of reflected ray is

5x+12y=13

Point of intersection of these line is (2,41​)

Now the normal at this point will be the bisector to the angle between these lines .

So, the equation of normal is 53x−4y−5​=±135x+12y−13​

⇒39x−52y−65=±(25x+60y−65)

So, we get the equation of normals as

x=8y and 32x+4y−65=0

So, the equation of lines are x=8y and 32x+4y−65=0

solution