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

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

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Answer 1 · 2017-09-13T16:50:22

$a + b = 60$ $a+b = 60$

Explanation:

We are given 3 equations:

$x - 2 y - 3 = 0 [1]$ $x-2y-3=0" [1]"$
$3 x - 2 y - 5 = 0 [2]$ $3x-2y-5=0" [2]"$
$a x - 2 y = b [3]$ $ax-2y = b" [3]"$

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

$- 2 y = - x + 3$ $-2y = -x+3$

$y = \frac{1}{2} x - \frac{3}{2}$ $y = 1/2x-3/2$

$m_{1} = \frac{1}{2}$ $m_1 = 1/2$

The angle that it forms with the x axis is $θ_{1} = {tan}^{- 1} (\frac{1}{2})$ $theta_1 = tan^-1(1/2)$

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

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

$y = \frac{3}{2} x - \frac{5}{2}$ $y=3/2x-5/2$

$m_{2} = \frac{3}{2}$ $m_2=3/2$

The angle that it forms with the x axis is $θ_{2} = {tan}^{- 1} (\frac{3}{2})$ $theta_2 = tan^-1(3/2)$

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

$θ_{i} = {tan}^{- 1} (\frac{3}{2}) - {tan}^{- 1} (\frac{1}{2}) [4]$ $theta_i = tan^-1(3/2)-tan^-1(1/2)" [4]"$

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

$- 2 y = - a x + b$ $-2y = -ax+ b$

$y = \frac{a}{2} x - \frac{b}{2}$ $y = a/2x-b/2$

$m_{3} = \frac{a}{2}$ $m_3 = a/2$

The angle that it forms with the x axis is $θ_{3} = {tan}^{- 1} (\frac{a}{2})$ $theta_3 = tan^-1(a/2)$

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

$θ_{r} = {tan}^{- 1} (\frac{a}{2}) - {tan}^{- 1} (\frac{3}{2}) [5]$ $theta_r = tan^-1(a/2) - tan^-1(3/2)" [5]"$

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:

${tan}^{- 1} (\frac{a}{2}) - {tan}^{- 1} (\frac{3}{2}) = {tan}^{- 1} (\frac{3}{2}) - {tan}^{- 1} (\frac{1}{2})$ $tan^-1(a/2) - tan^-1(3/2) = tan^-1(3/2)-tan^-1(1/2)$

${tan}^{- 1} (\frac{a}{2}) = 2 {tan}^{- 1} (\frac{3}{2}) - {tan}^{- 1} (\frac{1}{2})$ $tan^-1(a/2) = 2tan^-1(3/2)-tan^-1(1/2)$

$a = 2 tan (2 {tan}^{- 1} (\frac{3}{2}) - {tan}^{- 1} (\frac{1}{2}))$ $a = 2tan(2tan^-1(3/2)-tan^-1(1/2))$

$a = 29$ $a = 29$

Substitute into equation [3]:

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

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

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

Subtract [1] from [2]:

$2 x - 2 = 0$ $2x -2 = 0$

$x = 1$ $x = 1$

Substitute 1 for x into equation [1]:

$1 - 2 y - 3 = 0$ $1-2y-3=0$

#-2y -2 = 0

$y = - 1$ $y = -1$

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

Equation [3.1] must contain the same point:

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

$b = 31$ $b = 31$

$a + b = 60$ $a+b = 60$

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

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Explanation:

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

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Explanation:

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