Question

A ray of light is sent along the line x - 2y = 3 and upon reaching the line 3x-2y=5, the ray is reflected from it. Find the equation of line containing the reflected ray.​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

A ray of light is sent along the line x - 2y = 3 and upon reaching the line 3x-2y=5, the ray is reflected from it.

So, incident ray lies along the line x - 2y = 3 and the line 3x - 2y = 5 acts as a mirror.

Let assume that P be the point where incident ray strike the mirror.

So, solving two lines to get coordinates of P.

$\rm \: x - 2y = 3 - - - (1)$

and

$\rm \: 3x - 2y = 5 - - - (2)$

On Subtracting equation (1) from (2), we get

$\rm \: 2x = 2$

$\rm\implies \:x = 1$

So, on substituting x = 1 in equation (1), we get

$\rm \: 1 - 2y = 3$

$\rm \: - 2y = 3 - 1$

$\rm \: - 2y = 2$

$\rm\implies \:y = - 1$

So, it means Coordinates of P (1, - 1).

Now, from figure [ in attachment ] we concluded that

∠APB = ∠CPD = a ( which is acute ).

Now,

Equation of incident ray is x - 2y = 3

So, slope of incident ray, m = 1/2

Also,

Equation of mirror is 3x - 2y = 5

So, slope of mirror, M = 3/2

Let assume that slope of line along reflected ray be n.

We know,

The angle p between two lines slope m and M is given by

$\boxed{ \rm{ \:\rm \: tanp = \bigg |\dfrac{m - M}{1 + mM} \bigg| \: }}$

So, on substituting the values, we get

$\rm \: tana = \bigg |\dfrac{\dfrac{1}{2} - \dfrac{3}{2} }{1 + \dfrac{1}{2} \times \dfrac{3}{2} } \bigg| = \bigg |\dfrac{n - \dfrac{3}{2} }{1 + n \times \dfrac{3}{2} } \bigg|$

$\rm \: \bigg | \frac{ - 4}{7} \bigg| = \bigg | \frac{2n - 3}{2 + 3n} \bigg|$

$\rm \: \dfrac{4}{7} = \bigg | \frac{2n - 3}{2 + 3n} \bigg|$

$\rm \: \frac{2n - 3}{2 + 3n} = \: \pm \: \dfrac{4}{7}$

$\rm \: \frac{2n - 3}{2 + 3n} =\dfrac{4}{7} \: \: \: or \: \: \: \frac{2n - 3}{2 + 3n} = - \dfrac{4}{7}$

$\rm \: 14n - 21 = 8 + 12n \: \: or \: \: 14n - 21 = - 8 - 12n$

$\rm \: 2n = 29 \: \: or \: \: 26n = 13$

$\bf\implies \:n = \dfrac{29}{2} \: \: \: or \: \rm \: n = \dfrac{1}{2} \: \{rejected \: as \: its \: slope \: of \: incident \: ray\}$

So,

Equation of line containing the reflected ray which passes through the point (1, - 1) and having slope n = 29/2 is given by

$\rm \: y - ( - 1) = \dfrac{29}{2}(x - 1)$

$\rm \: 2y + 2 = 29x - 29$

$\rm \: 29x - 2y - 31 = 0$

So, the equation of line containing reflected ray is

$\rm\implies \:\boxed{ \rm{ \:\rm \: 29x - 2y - 31 = 0 \: \: }}$

$\rule{190pt}{2pt}$

Additional Information :-

Different forms of equations of a straight line

1. Equations of horizontal and vertical lines

Equation of line parallel to y - axis passes through the point (a, b) is x = a.

Equation of line parallel to x - axis passes through the point (a, b) is y = b.

2. Point-slope form equation of line

Equation of line passing through the point (a, b) having slope m is y - b = m(x - a)

3. Slope-intercept form equation of line

Equation of line which makes an intercept of c units on y axis and having slope m is y = mx + c.

4. Intercept Form of Line

Equation of line which makes an intercept of a and b units on x - axis and y - axis respectively is x/a + y/b = 1.

5. Normal form of Line

Equation of line which is at a distance of p units from the origin and perpendicular makes an angle β with the positive X-axis is x cosβ + y sinβ = p.