Why do we assume the parallel line is $4x - 3y + \lambda = 0$?

Great question! Parallel lines maintain the same slope. Since the slope of $Ax + By + C = 0$ is $-A/B$, any line with the same slope must have the same ratio of $A/B$. Thus, $4x - 3y$ remains constant, and only the constant term $\lambda$ varies.

JEE Main 2019 (10 April Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Straight Lines: 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 ?

Select Answer:

Progressive Hints

Hint 1

The Spark: When you see the phrase 'parallel to the line', your mind should immediately jump to the slope-intercept form or the general form. If two lines are parallel, they share the same slope. This means the coefficients of

x

and

y

remain identical, while only the constant term changes. Visualize this as shifting the line up or down the plane without tilting it.

Hint 2

The Arsenal: To find the unknown constant, you need the perpendicular distance formula from the origin

(0, 0)

to a line

A x + B y + C = 0

, which is

d = \frac{∣ C ∣}{A ^{2} + B ^{2}}

. This is your primary tool for bridging the gap between the geometric constraint and the algebraic equation.

Hint 3

The Execution: Start by writing the general equation of the parallel lines as

4 x - 3 y + λ = 0

. Now, apply the distance formula:

\frac{∣ λ ∣}{4 ^{2} + ( - 3 ) ^{2}} = \frac{3}{5}

. Solve this equation for

λ

, keeping in mind that the absolute value will yield two distinct values, representing the two lines on either side of the origin.

Visualized Solution

The Reference Line

Given reference line: $4 x - 3 y + 2 = 0$
We need to find lines parallel to this reference line.

Equation of Parallel Lines

Any line parallel to $A x + B y + C = 0$ has the form $A x + B y + λ = 0$ .
So, our parallel lines will be: $4 x - 3 y + λ = 0$

The Distance Constraint

The problem states these parallel lines are at a distance of $\frac{3}{5}$ from the origin $(0, 0)$ .

Distance Formula from Origin

The perpendicular distance $d$ from $(0, 0)$ to a line $A x + B y + C = 0$ is given by:
$d = \frac{∣ C ∣}{A ^{2} + B ^{2}}$

Substituting the Values

Substitute $A = 4$ , $B = - 3$ , $C = λ$ , and $d = \frac{3}{5}$ :
$\frac{∣ λ ∣}{( 4 ) ^{2} + ( - 3 ) ^{2}} = \frac{3}{5}$

Simplifying the Denominator

Calculate the denominator: $16 + 9 = 25 = 5$
The equation becomes: $\frac{∣ λ ∣}{5} = \frac{3}{5}$

Solving for $λ$

Multiply both sides by $5$ : $∣ λ ∣ = 3$
Removing the absolute value gives two possibilities:
$λ = 3$ or $λ = - 3$

The Two Parallel Lines

Substituting $λ$ back, we get two lines:
$L_{1} : 4 x - 3 y + 3 = 0$
$L_{2} : 4 x - 3 y - 3 = 0$

Testing the Given Options

We need to find which of the given points lies on either $L_{1}$ or $L_{2}$ .
Let's test the point $(- \frac{1}{4}, \frac{2}{3})$ .

Substituting the Point in $L_{1}$

Substitute $x = - \frac{1}{4}$ and $y = \frac{2}{3}$ into $L_{1} : 4 x - 3 y + 3 = 0$
$4 (- \frac{1}{4}) - 3 (\frac{2}{3}) + 3 = 0$

Evaluating the Expression

$4 (- \frac{1}{4}) = - 1$
$- 3 (\frac{2}{3}) = - 2$
$- 1 - 2 + 3 = - 3 + 3 = 0$

Final Conclusion

Since the result is $0$ , the point $(- \frac{1}{4}, \frac{2}{3})$ satisfies the equation of $L_{1}$ .
Therefore, this point lies on one of the parallel lines.

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The Sigma Insight: Distance of a Point from a Line

Solution Diagram

Analyzing the Setup

When we say two lines are parallel, we are essentially saying they share the same "tilt" or slope. In the general equation of a line,

A x + B y + C = 0

, the slope is determined by the ratio of the coefficients of

x

and

y

If two lines are parallel, they must have the same slope, which means their

A

and

B

coefficients must be proportional. For simplicity, we can keep them identical.

Thus, any line parallel to

4 x - 3 y + 2 = 0

can be written in the form:

4 x - 3 y + λ = 0

where

λ

is an unknown constant. This

λ

is the "magic number" that shifts our line across the plane.

The Distance Constraint

The problem states that these parallel lines are at a distance of

3/5

units from the origin

(0, 0)

. The perpendicular distance

d

from the origin to a line

A x + B y + C = 0

is given by the formula:

d = \frac{∣ C ∣}{A ^{2} + B ^{2}}

By substituting our values,

A = 4

B = - 3

, and

C = λ

, we obtain:

\frac{∣ λ ∣}{4 ^{2} + ( - 3 ) ^{2}} = \frac{3}{5}

Solving for the Constant

The denominator,

16 + 9

, simplifies beautifully to

25

, which is

5

. Our equation becomes:

\frac{∣ λ ∣}{5} = \frac{3}{5}

Multiplying both sides by

5

, we find that

∣ λ ∣ = 3

. Because of the absolute value,

λ

can be either

3

- 3

This confirms our geometric intuition: there are two lines, one on each side of the origin, that satisfy the condition. Our two lines are:

4 x - 3 y + 3 = 0 and 4 x - 3 y - 3 = 0

Final Verification

To conclude, we test the given point

(- \frac{1}{4}, \frac{2}{3})

in the first equation:

4 (- \frac{1}{4}) - 3 (\frac{2}{3}) + 3 = - 1 - 2 + 3 = 0

The point satisfies the equation perfectly. Through these steps, we have navigated the plane to confirm the correct line equation.

Common Traps & FAQs

Why do we assume the parallel line is

4 x - 3 y + λ = 0

Great question! Parallel lines maintain the same slope. Since the slope of

A x + B y + C = 0

- A / B

, any line with the same slope must have the same ratio of

A / B

. Thus,

4 x - 3 y

remains constant, and only the constant term

λ

varies.

Why is there an absolute value in the distance formula?

Distance is a physical quantity, and it must always be non-negative. The expression

A x + B y + C

can be positive or negative depending on which side of the line the origin lies. The absolute value ensures we are calculating the magnitude of the perpendicular gap, regardless of the sign.

Why are there two lines? Shouldn't there be just one?

Think of the origin as a reference point. A line at a fixed distance from the origin can exist on either side of it. Imagine a ruler placed at a distance of

3/5

units from the origin; you could place it above or below, resulting in two distinct parallel lines.

Animated Solution for Mathematics - Straight Lines: 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 ?

Select Answer:

Visualized Solution

The Reference Line

Equation of Parallel Lines

The Distance Constraint

Distance Formula from Origin

Substituting the Values

Simplifying the Denominator

Solving for $λ$

The Two Parallel Lines

Testing the Given Options

Substituting the Point in $L_{1}$

Evaluating the Expression

Final Conclusion

The Sigma Insight: Distance of a Point from a Line

Analyzing the Setup

The Distance Constraint

Solving for the Constant

Final Verification

Similar Questions

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Select Answer:

Visualized Solution

The Reference Line

Equation of Parallel Lines

The Distance Constraint

Distance Formula from Origin

Substituting the Values

Simplifying the Denominator

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

The Two Parallel Lines

Testing the Given Options

Substituting the Point in .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L1​

Evaluating the Expression

Final Conclusion

The Sigma Insight: Distance of a Point from a Line

Analyzing the Setup

The Distance Constraint

Solving for the Constant

Final Verification

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Substituting the Point in $L_{1}$

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