A ray of light coming from (2,23) is incidentat an
angle 30° on the line x = 1 at the point A. The ray
gets reflected on the line x = 1 and meets x axis at
a point B. Then the line AB passes through the point
Answers
Step-by-step explanation:
Given that,
There is a point B on the x−axis on which rays reflects
A ray passing through A(1,2) reflects on point B.
On reflection the ray passes through point C(5,3)
We need to find equation of line AB.
Since
the point B is on the x−axis
its y=0
Let the coordinates of point B(k,0)
Let the angle ∠CBX=θ
PB be the normal
∠PBX=90
o
∠XBC+∠CBP=90
o
θ+∠CBP=90
o
∠CBP=90
o
Also, ∠ABP=∠CBP=90
o
=θ
Now, ∠ABX=∠ABP+∠PBC+∠CBX
=(90
o
−θ)+(90
o
−θ)+θ
=180
o
−20+Q
∠ABX=180
o
−θ
Now, we find slope of line AB and CB
We know that
Slope of line
m=
x
2
−x
1
y
2
−y
1
Line AB
Slope of line AB passing through the points (1,2) and (k,0)
Slope of AB=
k−1
0−2
let m
1
=
k−1
−2
now,
∠ABx=180
o
−θ
m
1
=tan(180
o
−θ)
=−tanθ
m
1
=−tanQ=
k−1
−2
m
1
=tanQ=
k−1
2
Line CB Slope of line CB passing through the points (5,3) and (k,0)
Slope of CB=
k−5
0−3
m
2
=
k−5
−3
But m
2
=tanθ=
k−1
2
According to question
m
1
=m
2
k−1
2
=
k−5
−3
2k−10=−3k+3
2k+3k=3+10
5k=13
k=
5
13
Then point B(k,0)=B(
5
13
,0)=(x
2
,y
2
)
Equation of line AB
y−y
1
=
x
2
−x
1
y
2
−y
1
y−2=
5
13
−1
0−2
(x−1)
y−2=
5
13−5
−2
(x−1)
y−2=
8
−10
(x−1)
y−2=
4
−5
(x−1)
4y−8=−5x+5
5x+4y−13=0
Hence, the equation of line AB.
This is the answer.
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