JEE Main 2024 (09 Apr Shift 1)

LEVELJEE Advanced

Animated Solution for Mathematics - Straight Lines: A ray of light coming from the point $P (1, 2)$ gets reflected from the point $Q$ on the $x$ -axis and then passes through the point $R (4, 3)$ . If the point $S (h, k)$ is such that $P QR S$ is a parallelogram, then $h k^{2}$ is equal to :

Select Answer:

Progressive Hints

Hint 1

The first step to mastering this problem is to visualize the path of the light ray. When a ray of light reflects off a surface, it follows the law of reflection. Instead of calculating the bounce directly, imagine the mirror (the x-axis) as a portal. If you reflect the source point

P (1, 2)

across the x-axis to create a virtual image

P^{'}

, the path from

P^{'}

R

becomes a perfectly straight line. Can you find the coordinates of

P^{'}

Hint 2

Once you have the image point

P^{'} (1, - 2)

, recognize that the points

P^{'}

Q

, and

R

are collinear. This is the core geometric reality. Use the slope formula

m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}

to find the equation of the line passing through

P^{'}

and

R

. The point

Q

is simply the x-intercept of this line, where

y = 0

Hint 3

For the final part, you are dealing with a parallelogram

P QR S

. The most powerful property of a parallelogram is that its diagonals bisect each other. This means the midpoint of diagonal

P R

must be identical to the midpoint of diagonal

QS

. Set the midpoint of

P R

equal to the midpoint of

QS

to solve for the coordinates of

S (h, k)

Visualized Solution

Visualizing the Ray Reflection

Given points: $P (1, 2)$ and $R (4, 3)$ .
A light ray from $P$ reflects at $Q (x, 0)$ on the x-axis.
It then passes through $R$ .

Finding the Image Point $P^{'}$

By the laws of reflection, the reflected ray $QR$ appears to originate from the virtual image of $P$ .
The image of $P (1, 2)$ across the x-axis ( $y = 0$ ) is $P^{'} (1, - 2)$ .

Collinearity of $P^{'}$ , $Q$ , and $R$

The points $P^{'} (1, - 2)$ , $Q (x, 0)$ , and $R (4, 3)$ all lie on the same straight line.
We can find the slope of this line using $P^{'}$ and $R$ .
$m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}} = \frac{3 - ( - 2 )}{4 - 1} = \frac{5}{3}$ .

Equation of Line $P^{'} R$

Using the point-slope form: $y - y_{1} = m (x - x_{1})$ .
Substituting $R (4, 3)$ and $m = \frac{5}{3}$ : $y - 3 = \frac{5}{3} (x - 4)$ .
Simplifying: $3 y - 9 = 5 x - 20 ⟹ 5 x - 3 y = 11$ .

Locating Point $Q$ on the X-axis

Point $Q$ is the x-intercept of the line $P^{'} R$ .
Substitute $y = 0$ into $5 x - 3 y = 11$ .
$5 x - 0 = 11 ⟹ x = \frac{11}{5}$ .
Therefore, $Q$ is $(\frac{11}{5}, 0)$ .

Constructing Parallelogram $P QR S$

We are given a point $S (h, k)$ such that $P QR S$ forms a parallelogram.
Let's visualize this parallelogram connecting $P$ , $Q$ , $R$ , and $S$ .

Property of Parallelogram Diagonals

In any parallelogram, the diagonals bisect each other.
This means the midpoint of diagonal $P R$ must be exactly the same as the midpoint of diagonal $QS$ .

Calculating the Midpoint of $P R$

Let's find the midpoint $M$ of the diagonal connecting $P (1, 2)$ and $R (4, 3)$ .
$M = (\frac{1 + 4}{2}, \frac{2 + 3}{2}) = (\frac{5}{2}, \frac{5}{2})$ .

Equating Midpoints to Find $k$

The midpoint of $QS$ is $(\frac{\frac{11}{5} + h}{2}, \frac{0 + k}{2})$ .
Equating the y-coordinates: $\frac{k}{2} = \frac{5}{2}$ .
Solving for $k$ : $k = 5$ .

Equating Midpoints to Find $h$

Equating the x-coordinates: $\frac{\frac{11}{5} + h}{2} = \frac{5}{2}$ .
Multiplying by 2: $\frac{11}{5} + h = 5$ .
Solving for $h$ : $h = 5 - \frac{11}{5} = \frac{25 - 11}{5} = \frac{14}{5}$ .

Final Calculation of $h k^{2}$

We need to find the value of $h k^{2}$ .
Substitute $h = \frac{14}{5}$ and $k = 5$ .
$h k^{2} = (\frac{14}{5}) \times (5)^{2} = \frac{14}{5} \times 25$ .
$h k^{2} = 14 \times 5 = 70$ .

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The Sigma Insight: Various Forms of Equations of a Line

Solution Diagram

Analyzing the Reflection Principle

To solve for the point

Q

where the light ray strikes the mirror (the x-axis), we utilize the principle of the virtual image. By reflecting the source point

P (1, 2)

across the x-axis, we obtain the virtual image

P^{'} (1, - 2)

The path from

P^{'}

to the target

R (4, 3)

forms a straight line. This transformation allows us to bypass complex trigonometric calculations involving angles of incidence and reflection.

Finding the Path of the Light

We determine the equation of the line passing through

P^{'} (1, - 2)

and

R (4, 3)

. The slope

m

is calculated as follows:

m = \frac{3 - ( - 2 )}{4 - 1} = \frac{5}{3}

Using the point-slope form

y - y_{1} = m (x - x_{1})

, we substitute the coordinates of

R

y - 3 = \frac{5}{3} (x - 4)

3 y - 9 = 5 x - 20

5 x - 3 y = 11

Since point

Q

lies on the x-axis, its y-coordinate is

0

. Substituting

y = 0

into the equation

5 x - 3 y = 11

, we get

5 x = 11

, which yields

x = \frac{11}{5}

. Thus, the coordinates of

Q

are

(\frac{11}{5}, 0)

Utilizing Parallelogram Symmetry

A parallelogram

P QR S

is defined by the property that its diagonals bisect each other. This means the midpoint of diagonal

P R

must be identical to the midpoint of diagonal

QS

First, we calculate the midpoint

M

P R

using

P (1, 2)

and

R (4, 3)

M = (\frac{1 + 4}{2}, \frac{2 + 3}{2}) = (\frac{5}{2}, \frac{5}{2})

Let the coordinates of

S

(h, k)

. Since the midpoint of

QS

must also be

(\frac{5}{2}, \frac{5}{2})

, we set up the following equations:

\frac{0 + k}{2} = \frac{5}{2} \Rightarrow k = 5

\frac{\frac{11}{5} + h}{2} = \frac{5}{2} \Rightarrow \frac{11}{5} + h = 5

Solving for

h

h = 5 - \frac{11}{5} = \frac{25 - 11}{5} = \frac{14}{5}

Final Calculation

We have determined the coordinates of

S

to be

(\frac{14}{5}, 5)

. We are tasked with finding the value of

h k^{2}

h k^{2} = (\frac{14}{5}) \times (5)^{2}

h k^{2} = \frac{14}{5} \times 25 = 14 \times 5

The final result is 70.

Common Traps & FAQs

Wait, why do we reflect point P instead of point R? Does it change the answer?

That is a brilliant question! You could absolutely reflect point R across the x-axis to get

R^{'} (4, - 3)

instead. If you did that, the path

P Q R^{'}

would be a straight line. Both methods are mathematically equivalent and will lead you to the exact same point

Q

on the x-axis. The choice is yours, but the logic remains identical: we are turning a 'bent' path into a 'straight' path to make the geometry manageable.

Does the order of the vertices in the parallelogram PQRS matter?

It matters immensely! In geometry, when we name a polygon like

P QR S

, the vertices are listed in consecutive order around the perimeter. This means the sides are

P Q

QR

R S

, and

S P

. Consequently, the diagonals are

P R

and

QS

. If you assumed

P Q

and

R S

were the diagonals, you would be solving for a completely different shape. Always sketch the vertices in order to identify the correct diagonals.

Why can't we just use the slope of PQ and the slope of RS to solve for h and k?

You certainly could, but it would be much more tedious. You would have to calculate the slope of

P Q

, then the slope of

R S

, and set them equal, then do the same for the other pair of sides. Using the property that diagonals bisect each other is the 'JEE way'—it is faster, more elegant, and significantly reduces the chance of making a silly algebraic error. Always look for the most efficient path!

Animated Solution for Mathematics - Straight Lines: A ray of light coming from the point $P (1, 2)$ gets reflected from the point $Q$ on the $x$ -axis and then passes through the point $R (4, 3)$ . If the point $S (h, k)$ is such that $P QR S$ is a parallelogram, then $h k^{2}$ is equal to :

Select Answer:

Visualized Solution

Visualizing the Ray Reflection

Finding the Image Point $P^{'}$

Collinearity of $P^{'}$ , $Q$ , and $R$

Equation of Line $P^{'} R$

Locating Point $Q$ on the X-axis

Constructing Parallelogram $P QR S$

Property of Parallelogram Diagonals

Calculating the Midpoint of $P R$

Equating Midpoints to Find $k$

Equating Midpoints to Find $h$

Final Calculation of $h k^{2}$

The Sigma Insight: Various Forms of Equations of a Line

Analyzing the Reflection Principle

Finding the Path of the Light

Utilizing Parallelogram Symmetry

Final Calculation

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Select Answer:

Visualized Solution

Visualizing the Ray Reflection

Finding the Image Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P′

Equation of Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P′R

Locating Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q on the X-axis

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

Property of Parallelogram Diagonals

Calculating the Midpoint of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PR

Equating Midpoints to Find .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k

Equating Midpoints to Find .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h

Final Calculation of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } hk2

The Sigma Insight: Various Forms of Equations of a Line

Analyzing the Reflection Principle

Finding the Path of the Light

Utilizing Parallelogram Symmetry

Final Calculation

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