If x² - y² + 2hxy + 2gx + 2fy + c = 0 is the locus of a point, which moves such that it is always eq

Q: Why do we have to use the absolute value in the distance formula?

Distance is inherently a non-negative quantity. The absolute value ensures that regardless of which side of the line the point lies on, the distance remains positive. Without it, you would be calculating a signed distance, which would lead to incorrect results and miss the geometric reality of the locus.

Q: Is there a faster way than squaring and expanding?

You could use the direct angle bisector formula $\frac{a_1x + b_1y + c_1}{\sqrt{a_1^2 + b_1^2}} = \pm \frac{a_2x + b_2y + c_2}{\sqrt{a_2^2 + b_2^2}}$, but squaring is often safer to avoid sign errors. It also naturally leads to the joint equation of the pair of lines, which is exactly what the problem demands.

Q: Why did we divide by -3 at the end?

The problem provides a specific template: $x^2 - y^2 + 2hxy + 2gx + 2fy + c = 0$. Our derived equation started with $-3x^2$. To compare coefficients, we must normalize the equation so the coefficient of $x^2$ is $1$. Division is the key to unlocking the values of $h, g, f,$ and $c$.

Question

If $x^2 - y^2 + 2hxy + 2gx + 2fy + c = 0$ is the locus of a point, which moves such that it is always equidistant from the lines $x + 2y + 7 = 0$ and $2x - y + 8 = 0$, then the value of $g + c + h - f$ equals

Options:

(A) 14

(B) 6

(C) 8

(D) 29

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing at the intersection of two roads, represented by the lines $L_1: x + 2y + 7 = 0$ and $L_2: 2x - y + 8 = 0$. You are tasked with finding the path of a point $P(x, y)$ that maintains a perfect, unwavering balance—it must always be equidistant from both roads.

This is not just an algebraic exercise; it is the definition of an angle bisector. When you solve for this locus, you are essentially finding the lines that slice the angles between these two roads exactly in half.

### The Algebraic Compass

To translate this geometric intuition into the language of mathematics, we rely on the perpendicular distance formula. For any point $(x, y)$ and a line $Ax + By + C = 0$, the distance $d$ is given by:

$$d = \frac{|Ax + By + C|}{\sqrt{A^2 + B^2}}$$

We set the distance to $L_1$ equal to the distance to $L_2$. Substituting our lines, we get:

$$\frac{|x + 2y + 7|}{\sqrt{1^2 + 2^2}} = \frac{|2x - y + 8|}{\sqrt{2^2 + (-1)^2}}$$

Notice the elegance here: both denominators simplify to $\sqrt{5}$. This is no coincidence; it reveals that our original lines are perpendicular to each other! With the denominators canceling out, we are left with the beautiful simplicity of:

$$|x + 2y + 7| = |2x - y + 8|$$

### The Power of Difference of Squares

Now, we face the absolute value bars. A common mistake is to panic and expand everything immediately. Instead, let us be strategic.

By squaring both sides, we obtain $(x + 2y + 7)^2 = (2x - y + 8)^2$. Rather than expanding these trinomials, we bring everything to one side:

$$(x + 2y + 7)^2 - (2x - y + 8)^2 = 0$$

This is the classic difference of squares identity: $a^2 - b^2 = (a - b)(a + b)$. Applying this, we get:

$$[(x + 2y + 7) - (2x - y + 8)] \cdot [(x + 2y + 7) + (2x - y + 8)] = 0$$

Simplifying the terms inside the brackets yields $(-x + 3y - 1)(3x + y + 15) = 0$. These two linear factors are the equations of our two angle bisectors.

### The Final Normalization

To find the joint equation, we multiply these factors:

$$-3x^2 + 3y^2 + 8xy - 18x + 44y - 15 = 0$$

The problem asks us to match this with the form $x^2 - y^2 + 2hxy + 2gx + 2fy + c = 0$. Our equation starts with $-3x^2$, so we divide the entire expression by $-3$ to normalize it:

$$x^2 - y^2 - \frac{8}{3}xy + 6x - \frac{44}{3}y + 5 = 0$$

Now, we simply compare coefficients: $2h = -\frac{8}{3} \Rightarrow h = -\frac{4}{3}$, $2g = 6 \Rightarrow g = 3$, $2f = -\frac{44}{3} \Rightarrow f = -\frac{22}{3}$, and $c = 5$.

Finally, calculating $g + c + h - f$ gives us:

$$3 + 5 + \left(-\frac{4}{3}\right) - \left(-\frac{22}{3}\right) = 8 + \frac{18}{3} = 8 + 6 = 14$$

We have arrived at the destination. The path was clear, the logic was sound, and the result is **14**.

Animated Solution for Mathematics - Straight Lines: If $x^{2} - y^{2} + 2 h x y + 2 g x + 2 f y + c = 0$ is the locus of a point, which moves such that it is always equidistant from the lines $x + 2 y + 7 = 0$ and $2 x - y + 8 = 0$ , then the value of $g + c + h - f$ equals

Select Answer:

Visualized Solution

The Geometric Setup

Equating Perpendicular Distances

Substituting the Equations

Simplifying the Equation

Rearranging as Difference of Squares

Factoring the Expression

Simplifying the Factors

Expanding to the Joint Equation

Standardizing the Equation

Comparing Coefficients

Final Evaluation

The Sigma Insight: Angle Between Two Lines

Analyzing the Setup

The Algebraic Compass

The Power of Difference of Squares

The Final Normalization

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