Question

A ray of light passing through the point B (1, 2) reflects on the x-axis at point A and the
reflected ray passed through the point C (5, 3). Find the coordinates of the point A.

Answer 1

A ray of light BA strike on x-axis at A and reflect in AC direction. see the figure.

now ,
according to law of Reflection
angle of incidence ( i) = angle of reflection ( r)

hence, i = r
90° - i = 90° - r
∅1 = ∅2

take both sides tan

tan∅1 = tan∅2

we also know
tan∅ = | m1 - m2| /| 1 + m1.m2 |
where m1 and m2 are the slopes of two lines which inclined each other ∅ angle .
so,
tan∅1 = | { 2/(1-x) - 0|/| 1 + {2/(1-x)×0|
tan∅2 = |{5/(3-x) -0 |/| 1 + {5/(3 - x)|

[ note :- slope of x- axis = 0 ]

tan∅1 = tan∅2

| 2/( 1 - x)| = | 5/( 3 - x |
now ,break modulus ,

2/( 1 - x) = 5/( 3 - x)

2( 3 - x) = 5(1 - x)
6 - 2x = 5 -5x
1 = -3x
x = -1/3

again,
2/( 1 - x) = -5/( 3- x)
6-2x = -5 + 5x
11 = 7x
x = 11/7

now we have two value of x
e.g x = 11/7 and -1/3
but x ≠ -1/3 becoz both sides of points A are in First quadrant .

hence, x = 11/7

Answer 2

Step-by-step explanation:

Answer:(13/5,0).

Explanation:Hey,

It's known to us that the both rays reflected and incident are equally inclined to the normal at p.

Suppose coordinates of p =(t,0)

✓✓ show the diagram.

If pR makes an angle x with the normal at p, then it forms angle(90+x), with positive x axis and (90-x) is formed by ps with the positive x axis.

Now , slope of PR= tan (x+90) =cot(-x)

And slope of PS = tan(90-x)=cotx.

ATQ,

Slope of PS+slope of PR=0

=>[(3-0)/(5-t)]+[(0-2)/(t-1)]=0

=>3(t-1)+(-2)(5-t)=0

=>t=13/5

Hence, coordinate of p is (13/5,0)