JEE Main 2021 (25 July Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Straight Lines: Let the equation of the pair of lines, $y = p x$ and $y = q x$ , can be written as $(y - p x) (y - q x) = 0$ . Then the equation of the pair of the angle bisectors of the lines $x^{2} - 4 x y - 5 y^{2} = 0$ is:

Q: Wait, why can't we just find the individual lines first and then find their bisectors?

You absolutely can! You could factorize $x^2 - 4xy - 5y^2 = 0$ into $(x - 5y)(x + y) = 0$, find the individual lines $x - 5y = 0$ and $x + y = 0$, and then use the angle bisector formula for a single line. However, the combined equation formula $\frac{x^2 - y^2}{a - b} = \frac{xy}{h}$ is much faster and less prone to calculation errors during the final step. It is a vital tool for competitive exams.

Select Answer:

Progressive Hints

Hint 1

Hint 1: The Spark. Recognize that the equation

x^{2} - 4 x y - 5 y^{2} = 0

is a homogeneous equation of the second degree. This tells us that the equation represents two straight lines passing through the origin. Your task is to find the equation of the angle bisectors of these two lines, which will also pass through the origin.

Hint 2

Hint 2: The Arsenal. You do not need to find the individual lines. Use the standard formula for the pair of angle bisectors of a homogeneous equation

a x^{2} + 2 h x y + b y^{2} = 0

, which is

\frac{x ^{2} - y ^{2}}{a - b} = \frac{x y}{h}

. This formula is a powerful shortcut in coordinate geometry.

Hint 3

Hint 3: The Execution. Compare your given equation

x^{2} - 4 x y - 5 y^{2} = 0

with the general form

a x^{2} + 2 h x y + b y^{2} = 0

to identify

a

h

, and

b

. Specifically,

a = 1

2 h = - 4

(so

h = - 2

), and

b = - 5

. Now, substitute these values into the formula

\frac{x ^{2} - y ^{2}}{a - b} = \frac{x y}{h}

and simplify.

Visualized Solution

Visualizing the Pair of Lines

Given equation: $x^{2} - 4 x y - 5 y^{2} = 0$
This represents a pair of straight lines passing through the origin.
We need to find the equation of the pair of angle bisectors.

The General Formula

The general equation of a pair of lines is $a x^{2} + 2 h x y + b y^{2} = 0$ .
The equation of the pair of angle bisectors is given by:
$\frac{x ^{2} - y ^{2}}{a - b} = \frac{x y}{h}$

Identifying Coefficients

Comparing $x^{2} - 4 x y - 5 y^{2} = 0$ with $a x^{2} + 2 h x y + b y^{2} = 0$ :
$a = 1$
$2 h = - 4 \Rightarrow h = - 2$
$b = - 5$

Substitution in the Formula

Substitute $a = 1$ , $b = - 5$ , and $h = - 2$ into the formula:
$\frac{x ^{2} - y ^{2}}{1 - ( - 5 )} = \frac{x y}{- 2}$

Simplifying the Denominator

Simplify the denominator on the left side:
$\frac{x ^{2} - y ^{2}}{6} = \frac{x y}{- 2}$

Cross-Multiplication

Multiply both sides by $6$ to eliminate the fraction on the left:
$x^{2} - y^{2} = 6 \cdot \frac{x y}{- 2}$
$x^{2} - y^{2} = - 3 x y$

Final Equation

Rearranging the terms to one side:
$x^{2} + 3 x y - y^{2} = 0$
Key Takeaway: The pair of angle bisectors is always perpendicular to each other (sum of coefficients of $x^{2}$ and $y^{2}$ is zero).

00:00 / 00:00

The Sigma Insight: Angle Between Two Lines

Solution Diagram

Analyzing the Setup

The given equation is

x^{2} - 4 x y - 5 y^{2} = 0

. This is a homogeneous equation of the second degree, which represents a pair of straight lines passing through the origin.

Think of these lines as a pair of scissors pivoted at the origin. Our objective is to determine the equation of the pair of lines that bisect the angles between them.

The Magic of the General Formula

While one could factorize the equation to find individual lines, we prefer the elegant approach. For any pair of lines given by the general equation

a x^{2} + 2 h x y + b y^{2} = 0

, the combined equation of the angle bisectors is given by:

\frac{x ^{2} - y ^{2}}{a - b} = \frac{x y}{h}

This formula is a powerful tool in your JEE arsenal. It allows you to bypass the calculation of individual lines and proceed directly to the result.

Step-by-Step Execution

First, we extract the coefficients by comparing

x^{2} - 4 x y - 5 y^{2} = 0

with the general form

a x^{2} + 2 h x y + b y^{2} = 0

. We identify the parameters as follows:

a = 1

2 h = - 4 \Rightarrow h = - 2

b = - 5

Substituting these values into our bisector formula, we obtain:

\frac{x ^{2} - y ^{2}}{1 - ( - 5 )} = \frac{x y}{- 2}

The Final Simplification

Now, we simplify the algebraic expression. The denominator on the left side becomes

1 + 5 = 6

\frac{x ^{2} - y ^{2}}{6} = \frac{x y}{- 2}

Multiplying both sides by

6

, we get:

x^{2} - y^{2} = 6 \cdot (\frac{x y}{- 2})

x^{2} - y^{2} = - 3 x y

Rearranging all terms to one side, we arrive at the final equation:

$x^{2} + 3 x y - y^{2} = 0$

The Geometric Soul

Observe the final equation

x^{2} + 3 x y - y^{2} = 0

. Note that the sum of the coefficients of

x^{2}

and

y^{2}

1 + (- 1) = 0

In coordinate geometry, if the sum of the coefficients of

x^{2}

and

y^{2}

is zero, the lines represented by the equation are perpendicular. This confirms the geometric truth that the angle bisectors of any two intersecting lines must always be perpendicular to each other.

Common Traps & FAQs

Wait, why can't we just find the individual lines first and then find their bisectors?

You absolutely can! You could factorize

x^{2} - 4 x y - 5 y^{2} = 0

into

(x - 5 y) (x + y) = 0

, find the individual lines

x - 5 y = 0

and

x + y = 0

, and then use the angle bisector formula for a single line. However, the combined equation formula

\frac{x ^{2} - y ^{2}}{a - b} = \frac{x y}{h}

is much faster and less prone to calculation errors during the final step. It is a vital tool for competitive exams.

Why are the angle bisectors always perpendicular to each other?

That is a brilliant observation! If you look at the final equation

x^{2} + 3 x y - y^{2} = 0

, the sum of the coefficients of

x^{2}

and

y^{2}

1 + (- 1) = 0

. In any homogeneous equation of the form

A x^{2} + 2 H x y + B y^{2} = 0

, if

A + B = 0

, the lines represented by the equation are perpendicular. Since the angle bisectors of any two lines are always at

9 0^{\circ}

to each other, this property will always hold true for their combined equation.

Is there any risk of mixing up the signs for

a

h

, and

b

Yes, that is the most common trap! Always write out the general form

a x^{2} + 2 h x y + b y^{2} = 0

and align it directly under your given equation. For example, if you have

- 4 x y

, remember that

2 h = - 4

, not

h = - 4

. Taking that extra second to write it down prevents the most common silly mistakes.

Animated Solution for Mathematics - Straight Lines: Let the equation of the pair of lines, $y = p x$ and $y = q x$ , can be written as $(y - p x) (y - q x) = 0$ . Then the equation of the pair of the angle bisectors of the lines $x^{2} - 4 x y - 5 y^{2} = 0$ is:

Select Answer:

Visualized Solution

Visualizing the Pair of Lines

The General Formula

Identifying Coefficients

Substitution in the Formula

Simplifying the Denominator

Cross-Multiplication

Final Equation

The Sigma Insight: Angle Between Two Lines

Analyzing the Setup

The Magic of the General Formula

Step-by-Step Execution

The Final Simplification

The Geometric Soul

Similar Questions

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Select Answer:

Visualized Solution

Visualizing the Pair of Lines

The General Formula

Identifying Coefficients

Substitution in the Formula

Simplifying the Denominator

Cross-Multiplication

Final Equation

The Sigma Insight: Angle Between Two Lines

Analyzing the Setup

The Magic of the General Formula

Step-by-Step Execution

The Final Simplification

The Geometric Soul

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