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Animated Solution for Mathematics - Straight Lines: If the pair of lines $a x^{2} + 2 (a + b) x y + b y^{2} = 0$ lie along diameters of a circle and divide the circle into four sectors such that the area of one of the sectors is thrice the area of another sector then

Q: Why can't we just use the slope method?

You absolutely can! You could find the slopes $m_1$ and $m_2$ using the quadratic formula on the given equation and then use $\tan \theta = |\frac{m_1 - m_2}{1 + m_1m_2}|$. However, using the standard formula for the pair of lines is much faster and less prone to algebraic errors during the calculation. It is a more elegant, direct path.

Q: Why do we square both sides?

We square both sides to eliminate the radical $\sqrt{H^2 - AB}$ and the absolute value $|A+B|$. It is a standard algebraic technique to convert a radical equation into a polynomial equation, which is much easier to solve. Just remember that squaring can sometimes introduce extraneous solutions, though in this specific geometry problem, it is perfectly safe.

Select Answer:

Visualized Solution

Visualizing the Circle and Diameters

Given pair of lines: $a x^{2} + 2 (a + b) x y + b y^{2} = 0$
These represent straight lines passing through the origin.
Since they are diameters, they must intersect at the center $(0, 0)$ .

Relating Area to Central Angle

Area of a sector is proportional to its central angle: $Area \propto Angle$
Let the adjacent angles between the diameters be $θ$ and $18 0^{\circ} - θ$ .

Setting Up the Area Ratio

Given: Area of one sector is thrice the other.
Therefore, the ratio of their angles is also $3$ .
Equation: $\frac{18 0 ^{\circ} - θ}{θ} = 3$

Solving for the Angle $θ$

$18 0^{\circ} - θ = 3 θ$
$4 θ = 18 0^{\circ}$
$⟹ θ = 4 5^{\circ}$

The Angle Between Pair of Lines Formula

For a general pair of lines $A x^{2} + 2 H x y + B y^{2} = 0$ :
The angle $θ$ is given by: $tan θ = \frac{2 H ^{2} - A B}{∣ A + B ∣}$

Extracting and Substituting Coefficients

Comparing with $a x^{2} + 2 (a + b) x y + b y^{2} = 0$ :
Here: $A = a$ , $H = a + b$ , and $B = b$
Substitute these and $θ = 4 5^{\circ}$ into the formula.

Evaluating the Tangent Function

$tan 4 5^{\circ} = \frac{2 ( a + b ) ^{2} - ab}{∣ a + b ∣}$
Since $tan 4 5^{\circ} = 1$ :
$1 = \frac{2 ( a + b ) ^{2} - ab}{∣ a + b ∣}$

Simplifying the Numerator

Expand the term inside the square root:
$(a + b)^{2} - ab = a^{2} + 2 ab + b^{2} - ab$
This simplifies to: $a^{2} + ab + b^{2}$
So, $1 = \frac{2 a ^{2} + ab + b ^{2}}{∣ a + b ∣}$

Squaring Both Sides

To remove the square root and absolute value, square both sides:
$1^{2} = \frac{4 ( a ^{2} + ab + b ^{2} )}{( a + b ) ^{2}}$
Cross-multiply: $(a + b)^{2} = 4 (a^{2} + ab + b^{2})$

Expanding the Equation

Expand the left side: $a^{2} + 2 ab + b^{2}$
Distribute the $4$ on the right side: $4 a^{2} + 4 ab + 4 b^{2}$
Equating them: $a^{2} + 2 ab + b^{2} = 4 a^{2} + 4 ab + 4 b^{2}$

Rearranging to Final Form

Bring all terms to one side:
$4 a^{2} - a^{2} + 4 ab - 2 ab + 4 b^{2} - b^{2} = 0$
Simplifying gives: $3 a^{2} + 2 ab + 3 b^{2} = 0$

Conclusion

Final Condition: $3 a^{2} + 2 ab + 3 b^{2} = 0$
This perfectly matches Option 4.

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The Sigma Insight: Angle Between Two Lines

Solution Diagram

Analyzing the Setup

The equation

a x^{2} + 2 (a + b) x y + b y^{2} = 0

represents a pair of straight lines passing through the origin. These lines act as diameters of a circle, intersecting at the center.

The problem states that the area of one sector formed by these lines is three times the area of another. Since the area of a sector is directly proportional to its central angle, the ratio of the angles must also be

3 : 1

The Geometric Insight

Let the angle between the two lines be

θ

. The adjacent angle formed on the straight line is

18 0^{\circ} - θ

Given the ratio

3 : 1

, we establish the following relationship:

\frac{18 0 ^{\circ} - θ}{θ} = 3

Solving for

θ

18 0^{\circ} - θ = 3 θ

4 θ = 18 0^{\circ} \Rightarrow θ = 4 5^{\circ}

The Algebraic Bridge

We use the standard formula for the angle

θ

between a pair of lines represented by

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

tan θ = \frac{2 H ^{2} - A B}{∣ A + B ∣}

Comparing this to our given equation

a x^{2} + 2 (a + b) x y + b y^{2} = 0

, we identify the coefficients:

A = a

H = a + b

, and

B = b

Substituting these into the tangent formula with

θ = 4 5^{\circ}

tan 4 5^{\circ} = \frac{2 ( a + b ) ^{2} - ab}{∣ a + b ∣}

Since

tan 4 5^{\circ} = 1

, we have:

1 = \frac{2 ( a + b ) ^{2} - ab}{∣ a + b ∣}

Final Calculation

To solve for the relationship between

a

and

b

, we square both sides of the equation:

1 = \frac{4 (( a + b ) ^{2} - ab )}{( a + b ) ^{2}}

Cross-multiplying yields:

(a + b)^{2} = 4 (a^{2} + 2 ab + b^{2} - ab)

Expanding both sides:

a^{2} + 2 ab + b^{2} = 4 (a^{2} + ab + b^{2})

a^{2} + 2 ab + b^{2} = 4 a^{2} + 4 ab + 4 b^{2}

Rearranging all terms to one side, we obtain the final condition:

$3 a^{2} + 2 ab + 3 b^{2} = 0$

Common Traps & FAQs

Wait, why is the angle

4 5^{\circ}

? Shouldn't it be

13 5^{\circ}

Great question! The formula for the angle between a pair of lines gives the acute angle. Since the sectors are adjacent on a straight line, their angles sum to

18 0^{\circ}

. If the ratio is

3 : 1

, the angles are

4 5^{\circ}

and

13 5^{\circ}

. We use

4 5^{\circ}

because the tangent function is positive and it represents the acute angle between the lines.

Why can't we just use the slope method?

You absolutely can! You could find the slopes

m_{1}

and

m_{2}

using the quadratic formula on the given equation and then use

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

. However, using the standard formula for the pair of lines is much faster and less prone to algebraic errors during the calculation. It is a more elegant, direct path.

Why do we square both sides?

We square both sides to eliminate the radical

H^{2} - A B

and the absolute value

∣ A + B ∣

. It is a standard algebraic technique to convert a radical equation into a polynomial equation, which is much easier to solve. Just remember that squaring can sometimes introduce extraneous solutions, though in this specific geometry problem, it is perfectly safe.

Animated Solution for Mathematics - Straight Lines: If the pair of lines $a x^{2} + 2 (a + b) x y + b y^{2} = 0$ lie along diameters of a circle and divide the circle into four sectors such that the area of one of the sectors is thrice the area of another sector then

Select Answer:

Visualized Solution

Visualizing the Circle and Diameters

Relating Area to Central Angle

Setting Up the Area Ratio

Solving for the Angle $θ$

The Angle Between Pair of Lines Formula

Extracting and Substituting Coefficients

Evaluating the Tangent Function

Simplifying the Numerator

Squaring Both Sides

Expanding the Equation

Rearranging to Final Form

Conclusion

The Sigma Insight: Angle Between Two Lines

Analyzing the Setup

The Geometric Insight

The Algebraic Bridge

Final Calculation

Similar Questions

The combined equation of the two lines $a x + b y + c = 0$ and $a^{'} x + b^{'} y + c^{'} = 0$ can be written as $(a x + b y + c) (a^{'} x + b^{'} y + c^{'}) = 0$ . The equation of the angle bisectors of the lines represented by the equation $2 x^{2} + x y - 3 y^{2} = 0$ is

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:

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

If one of the lines of $m y^{2} + (1 - m^{2}) x y - m x^{2} = 0$ is a bisector of the angle between the lines $x y = 0$ , then $m$ is

Area of the triangle formed by the line $x + y = 3$ and angle bisectors of the pair of straight lines $x^{2} - y^{2} + 2 y = 1$ is

The pair of lines represented by $3 a x^{2} + 5 x y + (a^{2} - 2) y^{2} = 0$ are perpendicular to each other for

Let $P QR$ be a right angled isosceles triangle, right angled at $P (2, 1)$ . If the equation of the line $QR$ is $2 x + y = 3$ , then the equation representing the pair of lines $P Q$ and $P R$ is

Two poles, AB of length $a$ metres and CD of length $a + b (b \neq = a)$ metres are erected at the same horizontal level with bases at B and D. If $B D = x$ and $tan ∠ A C B = \frac{1}{2}$ , then:

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

Select Answer:

Visualized Solution

Visualizing the Circle and Diameters

Relating Area to Central Angle

Setting Up the Area Ratio

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

The Angle Between Pair of Lines Formula

Extracting and Substituting Coefficients

Evaluating the Tangent Function

Simplifying the Numerator

Squaring Both Sides

Expanding the Equation

Rearranging to Final Form

Conclusion

The Sigma Insight: Angle Between Two Lines

Analyzing the Setup

The Geometric Insight

The Algebraic Bridge

Final Calculation

Similar Questions

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Solving for the Angle $θ$

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

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The pair of lines represented by $3 a x^{2} + 5 x y + (a^{2} - 2) y^{2} = 0$ are perpendicular to each other for

Let $P QR$ be a right angled isosceles triangle, right angled at $P (2, 1)$ . If the equation of the line $QR$ is $2 x + y = 3$ , then the equation representing the pair of lines $P Q$ and $P R$ is

Two poles, AB of length $a$ metres and CD of length $a + b (b \neq = a)$ metres are erected at the same horizontal level with bases at B and D. If $B D = x$ and $tan ∠ A C B = \frac{1}{2}$ , then:

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