JEE Main 2020 - 6 Sep (Evening)

LEVELJEE Main

Animated Solution for Mathematics - Straight Lines: Let $L$ denote the line in the $x y$ -plane with $x$ and $y$ intercepts as 3 and 1 respectively. Then the image of the point $(- 1, - 4)$ in this line is :

Q: Why is there a $-2$ in the formula? Can't we just use $-1$?

Great question! The $-1$ is used when finding the *foot of the perpendicular* (the point on the line). Since the image is on the *other side* of the line at the same distance, we need to travel twice the distance from the point to the line. Hence, the factor of $-2$.

Q: Is it necessary to convert the intercept form to the general form $ax + by + c = 0$?

Absolutely. The image formula relies on the coefficients $a, b,$ and $c$. While you could technically work with fractions, converting to the general form $x + 3y - 3 = 0$ eliminates the risk of arithmetic errors and makes the substitution much cleaner.

Select Answer:

Progressive Hints

Hint 1

Visualize the line

L

in the

x y

-plane. It cuts the

x

-axis at

(3, 0)

and the

y

-axis at

(0, 1)

. Before diving into algebra, sketch this. The reflection of point

P (- 1, - 4)

will be a point

P^{'}

such that the line

L

acts as the perpendicular bisector of the segment

P P^{'}

. This is the geometric heart of the problem.

Hint 2

You have two paths: the 'long road' (finding the equation of the perpendicular line, finding the intersection point, and using the midpoint formula) or the 'express lane' (the direct image formula). For JEE Advanced, speed is key. Recall the formula for the image

(x^{'}, y^{'})

of a point

(x_{1}, y_{1})

across the line

a x + b y + c = 0

Hint 3

Apply the formula:

\frac{x ^{'} - x _{1}}{a} = \frac{y ^{'} - y _{1}}{b} = - 2 \frac{a x _{1} + b y _{1} + c}{a ^{2} + b ^{2}}

. Substitute

x_{1} = - 1, y_{1} = - 4, a = 1, b = 3, c = - 3

into this expression. Solve for

x^{'}

and

y^{'}

independently.

Visualized Solution

Visualizing Line $L$

Given intercepts: $x$ -intercept $= 3$ , $y$ -intercept $= 1$ .
The line passes through $(3, 0)$ and $(0, 1)$ .

Intercept Form of a Line

Intercept form of a line: $\frac{x}{a} + \frac{y}{b} = 1$
where $a$ is the $x$ -intercept and $b$ is the $y$ -intercept.

Substituting Intercepts

Substituting $a = 3$ and $b = 1$ :
$\frac{x}{3} + \frac{y}{1} = 1$

Converting to General Form

Multiply by $3$ : $x + 3 y = 3$
General form: $x + 3 y - 3 = 0$
Coefficients: $a = 1, b = 3, c = - 3$

Locating Point $P (- 1, - 4)$

Point to reflect: $P (x_{1}, y_{1}) = (- 1, - 4)$

The Image Formula

Formula for image $(x^{'}, y^{'})$ of $(x_{1}, y_{1})$ in line $a x + b y + c = 0$ :
$\frac{x ^{'} - x _{1}}{a} = \frac{y ^{'} - y _{1}}{b} = - 2 \frac{a x _{1} + b y _{1} + c}{a ^{2} + b ^{2}}$

Substituting Values into Formula

$\frac{x ^{'} - ( - 1 )}{1} = \frac{y ^{'} - ( - 4 )}{3} = - 2 \frac{1 ( - 1 ) + 3 ( - 4 ) - 3}{1 ^{2} + 3 ^{2}}$

Calculating the Numerator

Numerator: $1 (- 1) + 3 (- 4) - 3 = - 1 - 12 - 3 = - 16$

Calculating the Denominator

Denominator: $a^{2} + b^{2} = 1^{2} + 3^{2} = 1 + 9 = 10$

Finding the Common Ratio

Ratio $= - 2 \frac{- 16}{10} = \frac{32}{10} = \frac{16}{5}$

Solving for $x^{'}$

$\frac{x ^{'} + 1}{1} = \frac{16}{5}$
$x^{'} = \frac{16}{5} - 1 = \frac{11}{5}$

Solving for $y^{'}$

$\frac{y ^{'} + 4}{3} = \frac{16}{5}$
$y^{'} + 4 = \frac{48}{5}$
$y^{'} = \frac{48}{5} - 4 = \frac{28}{5}$

Final Image Coordinates

The image of $(- 1, - 4)$ is $(\frac{11}{5}, \frac{28}{5})$ .
Key Takeaway: The image formula is a direct shortcut for reflection problems in 2D geometry.

00:00 / 00:00

The Sigma Insight: Distance of a Point from a Line

Solution Diagram

Analyzing the Setup

The line

L

acts as a mirror for the point

P (- 1, - 4)

. We are given the

x

-intercept

a = 3

and the

y

-intercept

b = 1

Using the intercept form of a line,

\frac{x}{a} + \frac{y}{b} = 1

, we substitute the given values:

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

To simplify our calculations, we convert this into the general form

A x + B y + C = 0

. Multiplying the entire equation by

3

, we obtain:

x + 3 y - 3 = 0

Here, the coefficients are

A = 1

B = 3

, and

C = - 3

The Master Equation

To find the reflection

P^{'} (x^{'}, y^{'})

of point

P (x_{1}, y_{1})

across the line

A x + B y + C = 0

, we utilize the Image Formula:

\frac{x ^{'} - x _{1}}{A} = \frac{y ^{'} - y _{1}}{B} = - 2 \frac{A x _{1} + B y _{1} + C}{A ^{2} + B ^{2}}

This formula is highly efficient for competitive examinations as it bypasses the need for constructing perpendicular lines or calculating midpoints manually.

Final Calculation

First, we evaluate the expression for the point

P (- 1, - 4)

substituted into the line equation:

A x_{1} + B y_{1} + C = 1 (- 1) + 3 (- 4) - 3 = - 1 - 12 - 3 = - 16

Next, we calculate the denominator

A^{2} + B^{2}

A^{2} + B^{2} = 1^{2} + 3^{2} = 10

Now, we compute the constant ratio:

- 2 (\frac{- 16}{10}) = \frac{32}{10} = \frac{16}{5}

We solve for

x^{'}

and

y^{'}

individually:

For

x^{'}

\frac{x ^{'} - ( - 1 )}{1} = \frac{16}{5} \Rightarrow x^{'} = \frac{16}{5} - 1 = \frac{11}{5}

For

y^{'}

\frac{y ^{'} - ( - 4 )}{3} = \frac{16}{5} \Rightarrow y^{'} + 4 = \frac{48}{5} \Rightarrow y^{'} = \frac{48}{5} - 4 = \frac{28}{5}

The coordinates of the reflected point

P^{'}

are $(\frac{11}{5}, \frac{28}{5})$ .

Common Traps & FAQs

Why is there a

- 2

in the formula? Can't we just use

- 1

Great question! The

- 1

is used when finding the *foot of the perpendicular* (the point on the line). Since the image is on the *other side* of the line at the same distance, we need to travel twice the distance from the point to the line. Hence, the factor of

- 2

Is it necessary to convert the intercept form to the general form

a x + b y + c = 0

Absolutely. The image formula relies on the coefficients

a, b,

and

c

. While you could technically work with fractions, converting to the general form

x + 3 y - 3 = 0

eliminates the risk of arithmetic errors and makes the substitution much cleaner.

What if I forget the formula during the exam?

Don't panic! You can always derive it. Find the slope of the line

L

(

m = - 1/3

), then find the slope of the perpendicular line (

m^{'} = 3

). Write the equation of the line passing through

P (- 1, - 4)

with slope

3

. Find the intersection point

M

(the foot of the perpendicular). Finally, use the midpoint formula where

M

is the midpoint of

P P^{'}

. It takes longer, but it's a foolproof backup.

Animated Solution for Mathematics - Straight Lines: Let $L$ denote the line in the $x y$ -plane with $x$ and $y$ intercepts as 3 and 1 respectively. Then the image of the point $(- 1, - 4)$ in this line is :

Select Answer:

Visualized Solution

Visualizing Line $L$

Intercept Form of a Line

Substituting Intercepts

Converting to General Form

Locating Point $P (- 1, - 4)$

The Image Formula

Substituting Values into Formula

Calculating the Numerator

Calculating the Denominator

Finding the Common Ratio

Solving for $x^{'}$

Solving for $y^{'}$

Final Image Coordinates

The Sigma Insight: Distance of a Point from a Line

Analyzing the Setup

The Master Equation

Final Calculation

Similar Questions

The image of the point $(3, 5)$ in the line $x - y + 1 = 0$ , lies on :

If the image of the point $(- 4, 5)$ in the line $x + 2 y = 2$ lies on the circle $(x + 4)^{2} + (y - 3)^{2} = r^{2}$ , then $r$ is equal to:

Lines are drawn parallel to the line $4 x - 3 y + 2 = 0$ , at a distance $3/5$ from the origin. Then which one of the following points lies on any of these lines ?

Let the triangle PQR be the image of the triangle with vertices (1, 3), $(3, 1)$ and $(2, 4)$ in the line $x + 2 y = 2$ If the centroid of $△ P QR$ is the point $(α, β)$ , then $15 (α - β)$ is equal to:

A rectangle is formed by the lines $x = 0, y = 0, x = 3$ and $y = 4$ . Let the line L be perpendicular to $3 x + y + 6 = 0$ and divide the area of the rectangle into two equal parts. Then the distance of the point $(\frac{1}{2}, - 5)$ from the line $L$ is equal to :

Let the point $P (α, β)$ be at a unit distance from each of the two lines $L_{1} : 3 x - 4 y + 12 = 0$ , and $L_{2} : 8 x + 6 y + 11 = 0$ . If $P$ lies below $L_{1}$ and above $L_{2}$ , then $100 (α + β)$ is equal to

Line $L$ has intercepts $a$ and $b$ on the coordinate axes. When the axes are rotated through a given angle, keeping the origin fixed, the same line $L$ has intercepts $p$ and $q$ , then

Let $A (a, b), B (3, 4)$ and $(- 6, - 8)$ respectively denote the centroid, circumcentre and orthocentre of a triangle. Then, the distance of the point $P (2 a + 3, 7 b + 5)$ from the line $2 x + 3 y - 4 = 0$ measured parallel to the line $x - 2 y - 1 = 0$ is

The vertices of a triangle are $A (- 1, 3)$ , $B (- 2, 2)$ and $C (3, - 1)$ . A new triangle is formed by shifting the sides of the triangle by one unit inwards. Then the equation of the side of the new triangle nearest to origin is :

Select Answer:

Visualized Solution

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

Intercept Form of a Line

Substituting Intercepts

Converting to General Form

Locating Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(−1,−4)

The Image Formula

Substituting Values into Formula

Calculating the Numerator

Calculating the Denominator

Finding the Common Ratio

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

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

Final Image Coordinates

The Sigma Insight: Distance of a Point from a Line

Analyzing the Setup

The Master Equation

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The image of the point (3,5) in the line x−y+1=0, lies on :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the image of the point (−4,5) in the line x+2y=2 lies on the circle (x+4)2+(y−3)2=r2, then r is equal to:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Lines are drawn parallel to the line 4x−3y+2=0, at a distance 3/5 from the origin. Then which one of the following points lies on any of these lines ?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Line L has intercepts a and b on the coordinate axes. When the axes are rotated through a given angle, keeping the origin fixed, the same line L has intercepts p and q, then

Visualizing Line $L$

Locating Point $P (- 1, - 4)$

Solving for $x^{'}$

Solving for $y^{'}$

The image of the point $(3, 5)$ in the line $x - y + 1 = 0$ , lies on :

If the image of the point $(- 4, 5)$ in the line $x + 2 y = 2$ lies on the circle $(x + 4)^{2} + (y - 3)^{2} = r^{2}$ , then $r$ is equal to:

Lines are drawn parallel to the line $4 x - 3 y + 2 = 0$ , at a distance $3/5$ from the origin. Then which one of the following points lies on any of these lines ?

Line $L$ has intercepts $a$ and $b$ on the coordinate axes. When the axes are rotated through a given angle, keeping the origin fixed, the same line $L$ has intercepts $p$ and $q$ , then