Exercise Questions

All questions in this exercise are listed below. Click on a question to view its solution.

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Key Concepts

This exercise focuses on homogeneous second-degree equations representing pairs of straight lines through the origin.

Homogeneous Second-Degree Equation

A homogeneous equation of degree 2 in $x$ and $y$ has the form: $a{x}^2+2hxy+b{y}^2=0$

Every such equation (with real coefficients) represents two straight lines passing through the origin.

Decomposing into Separate Line Equations

To find the individual lines, treat the equation as a quadratic in $m=\frac{y}{x}$: $b{m}^2+2hm+a=0$ Solve for $m$ using the quadratic formula. Each value of $m$ gives a line $y=mx$ through the origin. Alternatively, factor the expression directly.

Nature of the Lines

The discriminant $h^2-ab$ determines the nature of the two lines:

Angle Between the Pair of Lines

The acute angle $\theta$ between the two lines is given by: $\tan \theta =\left|\frac{2\sqrt{h^2-ab}}{a+b}\right|$

• If $a+b=0$: the denominator is zero, so $\theta =90^{\circ }$ (lines are perpendicular).
• If $h^2=ab$: the numerator is zero, so $\theta =0^{\circ }$ (lines are coincident).

Condition for Perpendicular Lines

The two lines represented by $a{x}^2+2hxy+b{y}^2=0$ are perpendicular if and only if: $a+b=0$

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Important Formulas

General Form of Joint Equation: $a{x}^2+2hxy+b{y}^2=0$

Angle Between Lines: $\tan \theta =\left|\frac{2\sqrt{h^2-ab}}{a+b}\right|$

Conditions for Nature of Lines:

Perpendicularity Condition: $a+b=0$

Joint Equation of Perpendicular Lines:

If $a{x}^2+2hxy+b{y}^2=0$ represents a pair of lines, the joint equation of lines through the origin perpendicular to them is obtained by substituting $x\to y$, $y\to -x$: $b{x}^2-2hxy+a{y}^2=0$

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Worked Strategy

Step 1 — Identify $a$, $h$, $b$: Compare the given equation with $a{x}^2+2hxy+b{y}^2=0$. Remember the coefficient of $xy$ is $2h$, not $h$.

Step 2 — Check the discriminant $h^2-ab$: Determine whether the lines are real/distinct, coincident, or imaginary.

Step 3 — Find individual lines: Divide by $x^2$ and solve $b{\left(\frac{y}{x}\right)}^2+2h\left(\frac{y}{x}\right)+a=0$ for $m=y/x$.

Step 4 — Find the angle: Apply $\tan \theta =\left|\frac{2\sqrt{h^2-ab}}{a+b}\right|$.

Step 5 — Perpendicular joint equation: Swap $a\leftrightarrow b$ and negate $h$ to get $b{x}^2-2hxy+a{y}^2=0$.

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Summary

This exercise explores homogeneous second-degree equations $a{x}^2+2hxy+b{y}^2=0$ representing pairs of straight lines through the origin. The discriminant $h^2-ab$ classifies the lines as real/distinct, coincident, or imaginary. The angle between the lines uses the formula involving $\sqrt{h^2-ab}$ and $a+b$. Perpendicularity is confirmed by $a+b=0$. The joint equation of perpendicular lines is found by swapping the $x^2$ and $y^2$ coefficients and reversing the sign of the $xy$ term.