Question

A light beam emanating from the point a(3, 10) reflects from the straight line 2x + y -6 = 0 and then passes throughthe point b(4, 3). The equation of the reflected beam is

Answer 1

2X+Y-6 = 0 is the line as shown in the figure.  

A light beam is projected from (3,10) point onto this line and got reflected to point (4,3).

We need to find the equation of the reflected line as shown in figure.  

The angles of projection and reflection with the horizontal line are same.  

Let us assume (a,b) is the point on line 2X + Y – 6= 0 where the light is reflected.  

That means, the falling line (projected line) passing through points (3,10) and (a,b).

Substituting (a,b) in 2x + y – 6 = 0,  

we get b = -2a+6 ------------------------------E1.

Slope of the falling line = (y2 – y1)/(x2 – x1) = (b - 10) / (a - 3) = m

Equation of the falling line is  (line between (3,10) and (a,b))

(a – 3)y = (b – 10)x + (10a – 3b)  -------------------E2

Similarly slope of the reflected line is (y2 – y1)/(x2 – x1) = (b – 3)/(a – 4) = n

Equation of the reflected line is  (Line between (4, 3) and (a, b))

(a – 4)y = (b – 3)x + (3a – 4b) ---------------------------E3

Angle between 2 lines with slopes m1 and m2 is given by  

Tan(theta) = (m1 – m2)/(1 + m1*m2)

Since projected and reflected lines make same angle with 2x = y – 6 = 0, we can equate them  

First set of lines, m1 = -2  

m2 = (b-10)/(a – 3)                

From E1, we get m2 = (-2a – 4)/(a – 3) = -2(a -2)/(A – 3)

Tan(theta) = (2 – 2(a-2)/(a – 3))/(1 + 4(a – 2)/(a – 3))  

Simplifying we get,  

Tan(theta) = -2/(5a – 9)  ----------------------------E4

From second set of lines, m1 = -2,  

m2 = (b – 3)/(a – 4)  

From E1, we get m2 = (-2a + 6 – 3)/(a – 4) = -(2a – 3)/(a – 4)

Tan(theta) = (2 – (2a – 3)/(a – 4)) / (1 + 2(2a – 3)/(a – 4))

Simplifying we get,  

Tan(theta) = 1 / (a – 2) -----------------E5

Equating E4 and E5,  

-2/(5a – 9)  = 1 / (a – 2)

-2a + 4 = 5a – 9

7a = 13

a = 13/7

b = -2a + 6 = -2*13/7 + 6 =  6 – 26/7 = 16/7.

Both the lines touch 2x + y – 6 = 0 line at (13/7, 16/7)

Equation of reflected line is  

y(a – 4) = (b – 3)x + (3a – 4b)

(-15/7)y = (-5/7)x – (25/7)

Equation of reflected line is

5x - 15y + 25 = 0