A ray of light is sent along the line x-2y-3=0x2y3=0. Upon reaching the line3x-2y-5=03x2y5=0, the ray is reflected from it. If the equation of the line containing reflected ray is ax-2y = bax2y=b, then find the value of(a+b)(a+b)?

1 Answer
Sep 13, 2017

a+b = 60a+b=60

Explanation:

We are given 3 equations:

x-2y-3=0" [1]"x2y3=0 [1]
3x-2y-5=0" [2]"3x2y5=0 [2]
ax-2y = b" [3]"ax2y=b [3]

Find the slope of the line corresponding to equation [1]

-2y = -x+32y=x+3

y = 1/2x-3/2y=12x32

m_1 = 1/2m1=12

The angle that it forms with the x axis is theta_1 = tan^-1(1/2)θ1=tan1(12)

Find the slope of the line corresponding to equation [2]:

-2y=-3x+52y=3x+5

y=3/2x-5/2y=32x52

m_2=3/2m2=32

The angle that it forms with the x axis is theta_2 = tan^-1(3/2)θ2=tan1(32)

The angle from line [1] to line [2] is the angle of incidence:

theta_i = tan^-1(3/2)-tan^-1(1/2)" [4]"θi=tan1(32)tan1(12) [4]

Find the slope of the line corresponding to equation [3]:

-2y = -ax+ b2y=ax+b

y = a/2x-b/2y=a2xb2

m_3 = a/2m3=a2

The angle that it forms with the x axis is theta_3 = tan^-1(a/2)θ3=tan1(a2)

The angle from line [2] to line [3] is the angle of reflection:

theta_r = tan^-1(a/2) - tan^-1(3/2)" [5]"θr=tan1(a2)tan1(32) [5]

Because the angle of incidence equals the angle of reflection we can set the right side of equation [4] equal to the right side of equation 5:

tan^-1(a/2) - tan^-1(3/2) = tan^-1(3/2)-tan^-1(1/2)tan1(a2)tan1(32)=tan1(32)tan1(12)

tan^-1(a/2) = 2tan^-1(3/2)-tan^-1(1/2)tan1(a2)=2tan1(32)tan1(12)

a = 2tan(2tan^-1(3/2)-tan^-1(1/2))a=2tan(2tan1(32)tan1(12))

a = 29a=29

Substitute into equation [3]:

29x-2y = b" [3.1]"29x2y=b [3.1]

Find the point of intersection of lines [1] and [2]:

x-2y-3=0" [1]"x2y3=0 [1]
3x-2y-5=0" [2]"3x2y5=0 [2]

Subtract [1] from [2]:

2x -2 = 02x2=0

x = 1x=1

Substitute 1 for x into equation [1]:

1-2y-3=012y3=0

#-2y -2 = 0

y = -1y=1

The point is (1, -1)(1,1)

Equation [3.1] must contain the same point:

29(1) - 2(-1) = b29(1)2(1)=b

b = 31b=31

a+b = 60a+b=60