Understanding the Pair of Straight Lines Equation

A homogeneous equation of the second degree in x and y, such as $\text{Ax}^2 + 2\text{Bxy} + \text{Hy}^2 = 0$, represents a pair of straight lines passing through the origin (0,0).

Let the two individual lines represented by this equation be $\text{y} = \text{m}_1\text{x}$ and $\text{y} = \text{m}_2\text{x}$.

The combined equation of these two lines is $(\text{y} - \text{m}_1\text{x})(\text{y} - \text{m}_2\text{x}) = 0$.

Expanding this, we get:

$\text{y}^2 - \text{m}_2\text{xy} - \text{m}_1\text{xy} + \text{m}_1\text{m}_2\text{x}^2 = 0$

$\text{y}^2 - (\text{m}_1 + \text{m}_2)\text{xy} + \text{m}_1\text{m}_2\text{x}^2 = 0$

To match this with the given equation $\text{Ax}^2 + 2\text{Bxy} + \text{Hy}^2 = 0$, we need to rearrange our expanded equation in terms of descending powers of x or y or make the coefficients match by dividing by a suitable constant. Let's match coefficients by looking at the general form $\text{ax}^2 + 2\text{hxy} + \text{by}^2 = 0$.

Comparing $\text{m}_1\text{m}_2\text{x}^2 - (\text{m}_1 + \text{m}_2)\text{xy} + \text{y}^2 = 0$ with $\text{Ax}^2 + 2\text{Bxy} + \text{Hy}^2 = 0$, we can see the correspondence between the coefficients:

• Coefficient of $\text{x}^2$: $\text{A}$ corresponds to $\text{m}_1\text{m}_2$ (or proportional)
• Coefficient of $\text{xy}$: $2\text{B}$ corresponds to $-(\text{m}_1 + \text{m}_2)$ (or proportional)
• Coefficient of $\text{y}^2$: $\text{H}$ corresponds to $1$ (or proportional)

Let's divide the given equation by H (assuming H ≠ 0) to make the coefficient of $\text{y}^2$ equal to 1:

$\frac{\text{A}}{\text{H}}\text{x}^2 + \frac{2\text{B}}{\text{H}}\text{xy} + \text{y}^2 = 0$

Now, comparing $\text{m}_1\text{m}_2\text{x}^2 - (\text{m}_1 + \text{m}_2)\text{xy} + \text{y}^2 = 0$ with $\frac{\text{A}}{\text{H}}\text{x}^2 + \frac{2\text{B}}{\text{H}}\text{xy} + \text{y}^2 = 0$, we can equate the coefficients:

• Sum of slopes: $\text{m}_1 + \text{m}_2 = -\frac{2\text{B}}{\text{H}}$
• Product of slopes: $\text{m}_1\text{m}_2 = \frac{\text{A}}{\text{H}}$

Condition for Perpendicular Lines

Two lines with slopes $\text{m}_1$ and $\text{m}_2$ are perpendicular if the product of their slopes is -1, i.e., $\text{m}_1\text{m}_2 = -1$.

Using the relationship we found for the product of slopes from the given equation:

$\text{m}_1\text{m}_2 = \frac{\text{A}}{\text{H}}$

For the lines to be perpendicular, we must have:

$\frac{\text{A}}{\text{H}} = -1$

Multiplying both sides by H (assuming H ≠ 0):

$\text{A} = -\text{H}$

Rearranging the terms, we get the condition:

$\text{A} + \text{H} = 0$

If H = 0, the original equation becomes $\text{Ax}^2 + 2\text{Bxy} = 0$, which is $\text{x}(\text{Ax} + 2\text{By}) = 0$. This represents two lines: $\text{x} = 0$ (the y-axis) and $\text{Ax} + 2\text{By} = 0$. For these lines to be perpendicular, $\text{Ax} + 2\text{By} = 0$ must be the x-axis, which means A = 0. If A=0 and H=0, the condition A+H=0 holds. The equation becomes $2\text{Bxy} = 0$, representing the lines x=0 and y=0, which are perpendicular. Thus, the condition $\text{A} + \text{H} = 0$ holds universally for perpendicular lines represented by $\text{Ax}^2 + 2\text{Bxy} + \text{Hy}^2 = 0$.

Conclusion

The lines represented by the equation $\text{Ax}^2 + 2\text{Bxy} + \text{Hy}^2 = 0$ are perpendicular if the sum of the coefficient of $\text{x}^2$ and the coefficient of $\text{y}^2$ is equal to zero. In this case, the coefficients are A and H, so the condition for perpendicularity is $\text{A} + \text{H} = 0$.