JEE Main 2023 (29 January Shift 1)

LEVELJEE Advanced

Animated Solution for Mathematics - Straight Lines: A light ray emits from the origin making an angle $3 0^{\circ}$ with the positive x-axis. After getting reflected by the line $x + y = 1$ , if this ray intersects x-axis at Q, then the abscissa of Q is

Q: Why is the slope of the mirror $m = -1$? I always get confused with the negative sign.

Think of the equation $x + y = 1$ as $y = -x + 1$. In the slope-intercept form $y = mx + c$, the coefficient of $x$ is the slope. Here, that coefficient is $-1$. It represents a line that drops as it moves to the right, which makes perfect sense for a line connecting $(1,0)$ and $(0,1)$.

Q: I got $m_2 = \sqrt{3}$. How do I know if this is the correct slope for the reflected ray?

Great check! Since the mirror has a slope of $-1$ (an angle of $135^\circ$) and the incident ray has a slope of $1/\sqrt{3}$ (an angle of $30^\circ$), the reflected ray must be steeper. A slope of $\sqrt{3}$ corresponds to $60^\circ$, which is consistent with the physics of reflection. Trust your math!

Select Answer:

Progressive Hints

Hint 1

Hint 1: Visualize the geometry. You have a ray originating from the origin

(0, 0)

hitting a mirror line

x + y = 1

. Before you worry about the reflection, find the exact point of impact

P

by solving the system of linear equations formed by the ray and the mirror.

Hint 2

Hint 2: Recall the Law of Reflection in coordinate geometry. The angle between the incident ray and the mirror must equal the angle between the reflected ray and the mirror. Use the formula

tan θ = ∣ \frac{m _{2} - m _{1}}{1 + m _{1} m _{2}} ∣

where

m_{1}

is the slope of the incident ray,

m_{2}

is the slope of the reflected ray, and

m

is the slope of the mirror.

Hint 3

Hint 3: Once you have the slope

m_{2}

of the reflected ray and the point of impact

P (x_{P}, y_{P})

, write the equation of the reflected line using the point-slope form:

y - y_{P} = m_{2} (x - x_{P})

. To find the abscissa of

Q

, simply set

y = 0

and solve for

x

Visualized Solution

$C oor d ina t e S y s t e m & M i r r or$

Coordinate Axes: $x$ -axis and $y$ -axis.
Reflecting Line (Mirror): $x + y = 1$ .
Intercepts of the mirror: $(1, 0)$ and $(0, 1)$ .

$I n c i d e n tR a y E q u a t i o n$

Angle of incidence with $x$ -axis: $θ = 3 0^{\circ}$ .
Slope of incident ray: $m_{1} = tan 3 0^{\circ} = \frac{1}{3}$ .
Equation of incident ray: $y = \frac{x}{3}$ .

$I n t er sec t i o n P o in tP$

Point $P$ is the intersection of $y = \frac{x}{3}$ and $x + y = 1$ .
We need to solve these two equations simultaneously.

$S u b s t i t u t i o n f or P$

Substitute $y = \frac{x}{3}$ into $x + y = 1$ :
$x + \frac{x}{3} = 1$

$C oor d ina t eso f P$

$x (1 + \frac{1}{3}) = 1 \Rightarrow x (\frac{3 + 1}{3}) = 1$
Abscissa of $P$ : $x = \frac{3}{3 + 1}$
Ordinate of $P$ : $y = \frac{1}{3 + 1}$
Coordinates of $P$ : $(\frac{3}{3 + 1}, \frac{1}{3 + 1})$

$L a w o f R e f l ec t i o n$

Slope of incident ray: $m_{1} = \frac{1}{3}$ .
Slope of mirror line ( $x + y = 1$ ): $m = - 1$ .
Let the slope of the reflected ray be $m_{2}$ .
By the law of reflection: $tan θ = ∣ \frac{m _{2} - m}{1 + m _{2} m} ∣ = ∣ \frac{m - m _{1}}{1 + m m _{1}} ∣$

$S l o p e E q u a t i o n$

Substitute $m = - 1$ and $m_{1} = \frac{1}{3}$ :
$\frac{m _{2} - ( - 1 )}{1 + m _{2} ( - 1 )} = \frac{- 1 - \frac{1}{3}}{1 + ( - 1 ) ( \frac{1}{3} )}$
$\frac{m _{2} + 1}{1 - m _{2}} = \frac{- 3 - 1}{3 - 1}$

$C a l c u l a t in g m_{2}$

$(m_{2} + 1) (3 - 1) = (m_{2} - 1) (3 + 1)$
$m_{2} 3 - m_{2} + 3 - 1 = m_{2} 3 + m_{2} - 3 - 1$
$- 2 m_{2} = - 23 \Rightarrow m_{2} = 3$

$E q u a t i o n o f R e f l ec t e d R a y$

Using point-slope form at $P (\frac{3}{3 + 1}, \frac{1}{3 + 1})$ with $m_{2} = 3$ :
$y - \frac{1}{3 + 1} = 3 (x - \frac{3}{3 + 1})$

$F in d in g x - in t er ce ptQ$

To find the $x$ -intercept $Q$ , set $y = 0$ :
$0 - \frac{1}{3 + 1} = 3 x - \frac{3}{3 + 1}$
$3 x = \frac{3}{3 + 1} - \frac{1}{3 + 1} = \frac{2}{3 + 1}$

$F ina l R es u l t$

Solve for $x$ :
$x = \frac{2}{3 ( 3 + 1 )} = \frac{2}{3 + 3}$
Correct Option: $\frac{2}{3 + 3}$

00:00 / 00:00

The Sigma Insight: Angle Between Two Lines

Solution Diagram

Analyzing the Setup

Imagine standing at the origin of a coordinate plane. You hold a laser pointer, aiming it at an angle of

3 0^{\circ}

relative to the positive x-axis.

Before you lies a mirror, a simple line defined by the equation

x + y = 1

. This is a story of a photon embarking on a journey, hitting a barrier, and bouncing off toward a new destination.

The Point of Impact

Our first task is to find where the ray meets the mirror. We know the incident ray follows the line

y = m_{1} x

, where the slope is

m_{1} = tan 3 0^{\circ} = \frac{1}{3}

The mirror is the line

x + y = 1

. To find the intersection point

P

, we solve these simultaneously:

x + \frac{x}{3} = 1

Solving for

x

, we find the abscissa of

P

to be

x = \frac{3}{3 + 1}

. Consequently, the ordinate is

y = \frac{1}{3 + 1}

This point

P

is the 'collision site' where the physics of reflection takes over.

The Law of Reflection

Now, we must determine the path of the reflected ray. The Law of Reflection states that the angle of incidence equals the angle of reflection.

In coordinate geometry, we use the slope formula for the angle between two lines:

tan θ = \frac{m _{2} - m}{1 + m _{2} m} = \frac{m - m _{1}}{1 + m m _{1}}

Here,

m = - 1

is the slope of our mirror. By substituting

m = - 1

and

m_{1} = \frac{1}{3}

, we set up the equation:

\frac{m _{2} + 1}{1 - m _{2}} = \frac{- 3 - 1}{3 - 1}

After cross-multiplying and expanding both sides, the terms simplify beautifully to

- 2 m_{2} = - 23

, giving us

m_{2} = 3

. The reflected ray has a slope of

3

, which corresponds to an angle of

6 0^{\circ}

with the x-axis.

The Final Destination

With the slope

m_{2} = 3

and the point

P (\frac{3}{3 + 1}, \frac{1}{3 + 1})

in our toolkit, we write the equation of the reflected ray using the point-slope form:

y - \frac{1}{3 + 1} = 3 (x - \frac{3}{3 + 1})

We want to find where this ray strikes the x-axis at point

Q

. By setting

y = 0

, we isolate

x

0 - \frac{1}{3 + 1} = 3 x - \frac{3}{3 + 1}

Rearranging the terms, we get:

3 x = \frac{3}{3 + 1} - \frac{1}{3 + 1} = \frac{2}{3 + 1}

Finally, solving for

x

, we arrive at the final answer:

x = \frac{2}{3 ( 3 + 1 )} = \frac{2}{3 + 3}

Common Traps & FAQs

Wait, why do we use the tangent formula for the angle between lines? Can't we just use geometry?

You absolutely can! Using geometry with angles is elegant, but the tangent formula is the 'universal key' in coordinate geometry. It handles the signs automatically, ensuring you don't get lost in the orientation of the lines. It is a robust tool that works even when the geometry isn't immediately obvious.

Why is the slope of the mirror

m = - 1

? I always get confused with the negative sign.

Think of the equation

x + y = 1

y = - x + 1

. In the slope-intercept form

y = m x + c

, the coefficient of

x

is the slope. Here, that coefficient is

- 1

. It represents a line that drops as it moves to the right, which makes perfect sense for a line connecting

(1, 0)

and

(0, 1)

Is there a shortcut to finding the reflected ray without calculating the slope?

There is a method using the reflection of a point across a line, but it is often more algebraically intensive than finding the slope. For JEE Advanced, mastering the slope-angle relationship is more beneficial because it is applicable to a wider variety of problems involving mirrors and lenses.

Why did we set

y = 0

to find the abscissa of

Q

The abscissa is just a fancy word for the x-coordinate. Since the point

Q

lies on the x-axis, its y-coordinate must be zero by definition. Setting

y = 0

is the standard way to find the x-intercept of any line.

I got

m_{2} = 3

. How do I know if this is the correct slope for the reflected ray?

Great check! Since the mirror has a slope of

- 1

(an angle of

13 5^{\circ}

) and the incident ray has a slope of

1/ 3

(an angle of

3 0^{\circ}

), the reflected ray must be steeper. A slope of

3

corresponds to

6 0^{\circ}

, which is consistent with the physics of reflection. Trust your math!

Animated Solution for Mathematics - Straight Lines: A light ray emits from the origin making an angle $3 0^{\circ}$ with the positive x-axis. After getting reflected by the line $x + y = 1$ , if this ray intersects x-axis at Q, then the abscissa of Q is

Select Answer:

Visualized Solution

$C oor d ina t e S y s t e m & M i r r or$

$I n c i d e n tR a y E q u a t i o n$

$I n t er sec t i o n P o in tP$

$S u b s t i t u t i o n f or P$

$C oor d ina t eso f P$

$L a w o f R e f l ec t i o n$

$S l o p e E q u a t i o n$

$C a l c u l a t in g m_{2}$

$E q u a t i o n o f R e f l ec t e d R a y$

$F in d in g x - in t er ce ptQ$

$F ina l R es u l t$

The Sigma Insight: Angle Between Two Lines

Analyzing the Setup

The Point of Impact

The Law of Reflection

The Final Destination

Similar Questions

If the curve $x^{2} + 2 y^{2} = 2$ intersects the line $x + y = 1$ at two points $P$ and $Q$ , then the angle subtended by the line segment $P Q$ at the origin is:

A straight line $L$ through the point $(3, - 2)$ is inclined at an angle $6 0^{\circ}$ to the line $3 x + y = 1$ . If $L$ also intersects the x-axis, then the equation of $L$ is

If the two lines $x + (a - 1) y = 1$ and $2 x + a^{2} y = 1 (a \in R - {0, 1})$ are perpendicular, then the distance of their point of intersection from the origin is :-

The equation of one of the straight lines which passes through the point (1,3) and makes an angles $tan^{- 1} (2)$ with the straight line, $y + 1 = 32 x$ is

Let $A (1, 0), B (2, - 1)$ and $C (\frac{7}{3}, \frac{4}{3})$ be three points. If the equation of the bisector of the angle $A B C$ is $α x + β y = 5$ , then the value of $α^{2} + β^{2}$ is

If the line $l_{1} : 3 y - 2 x = 3$ is the angular bisector of the lines $l_{2} : x - y + 1 = 0$ and $l_{3} : α x + β y + 17 = 0$ , then $α^{2} + β^{2} - α - β$ is equal to

If the pair of straight lines $x^{2} - 2 p x y - y^{2} = 0$ and $x^{2} - 2 q x y - y^{2} = 0$ be such that each pair bisects the angle between the other pair, then

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } CoordinateSystem&Mirror

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } IncidentRayEquation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } IntersectionPointP

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } SubstitutionforP

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } CoordinatesofP

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } LawofReflection

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } SlopeEquation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculatingm2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } EquationofReflectedRay

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Findingx−interceptQ

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } FinalResult

The Sigma Insight: Angle Between Two Lines

Analyzing the Setup

The Point of Impact

The Law of Reflection

The Final Destination

Similar Questions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A straight line L through the point (3,−2) is inclined at an angle 60∘ to the line 3​x+y=1. If L also intersects the x-axis, then the equation of L is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the two lines x+(a−1)y=1 and 2x+a2y=1(a∈R−{0,1}) are perpendicular, then the distance of their point of intersection from the origin is :-

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The equation of one of the straight lines which passes through the point (1,3) and makes an angles tan−1(2​) with the straight line, y+1=32​x is

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the pair of straight lines x2−2pxy−y2=0 and x2−2qxy−y2=0 be such that each pair bisects the angle between the other pair, then

$C oor d ina t e S y s t e m & M i r r or$

$I n c i d e n tR a y E q u a t i o n$

$I n t er sec t i o n P o in tP$

$S u b s t i t u t i o n f or P$

$C oor d ina t eso f P$

$L a w o f R e f l ec t i o n$

$S l o p e E q u a t i o n$

$C a l c u l a t in g m_{2}$

$E q u a t i o n o f R e f l ec t e d R a y$

$F in d in g x - in t er ce ptQ$

$F ina l R es u l t$

If the curve $x^{2} + 2 y^{2} = 2$ intersects the line $x + y = 1$ at two points $P$ and $Q$ , then the angle subtended by the line segment $P Q$ at the origin is:

A straight line $L$ through the point $(3, - 2)$ is inclined at an angle $6 0^{\circ}$ to the line $3 x + y = 1$ . If $L$ also intersects the x-axis, then the equation of $L$ is

If the two lines $x + (a - 1) y = 1$ and $2 x + a^{2} y = 1 (a \in R - {0, 1})$ are perpendicular, then the distance of their point of intersection from the origin is :-

The equation of one of the straight lines which passes through the point (1,3) and makes an angles $tan^{- 1} (2)$ with the straight line, $y + 1 = 32 x$ is

Let $A (1, 0), B (2, - 1)$ and $C (\frac{7}{3}, \frac{4}{3})$ be three points. If the equation of the bisector of the angle $A B C$ is $α x + β y = 5$ , then the value of $α^{2} + β^{2}$ is

If the line $l_{1} : 3 y - 2 x = 3$ is the angular bisector of the lines $l_{2} : x - y + 1 = 0$ and $l_{3} : α x + β y + 17 = 0$ , then $α^{2} + β^{2} - α - β$ is equal to

If the pair of straight lines $x^{2} - 2 p x y - y^{2} = 0$ and $x^{2} - 2 q x y - y^{2} = 0$ be such that each pair bisects the angle between the other pair, then