Let a given line L₁ intersects the x and y axes at P and Q, respectively. Let another line L₂, perpe

Q: Wait, why is $c$ a variable? I thought $L_2$ was a specific line.

Great question! The problem states $L_2$ is perpendicular to $L_1$, but it doesn't say it passes through a specific point. It just says it cuts the axes. This means $L_2$ can slide along the plane while maintaining its slope. That 'sliding' is exactly what $c$ represents. It is a parameter that defines the family of all lines perpendicular to $L_1$.

Q: Why can't we just use the slope-point form for lines $PS$ and $QR$?

You absolutely can! However, the intercept form is often more elegant when dealing with axes intersections. It keeps the algebra symmetric and reduces the chance of sign errors. Always choose the path of least resistance in coordinate geometry; the intercept form is your best friend when you know where a line hits the axes.

Q: How do I know the final equation is a circle?

Look at the structure of the final result: $x^2 + y^2 - ax - by = 0$. A general circle equation is $x^2 + y^2 + 2gx + 2fy + c = 0$. Since our equation matches this form (with $c=0$ here), it is a circle passing through the origin. The presence of $x^2$ and $y^2$ with equal coefficients, and the absence of an $xy$ term, is the classic signature of a circle.

Question

Let a given line $L_1$ intersects the $x$ and $y$ axes at $P$ and $Q$, respectively. Let another line $L_2$, perpendicular to $L_1$, cut the $x$ and $y$ axes at $R$ and $S$, respectively. Show that the locus of the point of intersection of the lines $PS$ and $QR$ is a circle passing through the origin.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine standing before a blank coordinate plane. We draw our first line, $L_1$. It is a simple, elegant line that cuts the x-axis at $P(a, 0)$ and the y-axis at $Q(0, b)$.

Its equation, in the beautiful intercept form, is:

$$ \frac{x}{a} + \frac{y}{b} = 1 $$

From this, we immediately see its slope, $m_1 = -\frac{b}{a}$. Now, we introduce the second player, $L_2$. It is perpendicular to $L_1$.

In the world of geometry, perpendicularity is a powerful constraint. It dictates that the product of their slopes must be $-1$. Thus, the slope of $L_2$ must be $m_2 = \frac{a}{b}$.

Because $L_2$ is not fixed to a specific point, but is merely constrained by its slope, it can slide across the plane. We represent this freedom with a parameter $c$, giving us the equation:

$$ y = \frac{a}{b}x + c $$

This $c$ is our 'ghost' variable; it allows the line to move, and it is the very thing we will eventually exorcise from our equations.

### The Geometry of Intersection

With $L_2$ defined, we find its intercepts. Setting $x=0$ gives us the y-intercept $S(0, c)$. Setting $y=0$ gives us the x-intercept $R(-\frac{bc}{a}, 0)$.

Now, look at the four points we have: $P(a, 0)$, $Q(0, b)$, $R(-\frac{bc}{a}, 0)$, and $S(0, c)$. The problem asks us to consider the intersection of two new lines: $PS$ and $QR$.

Let their intersection point be $M(x, y)$. This point $M$ is the protagonist of our story. As $L_2$ slides (as $c$ changes), $M$ moves. We want to know the shape of the path $M$ traces.

### The Algebraic Exorcism

To find the locus, we need the equations of $PS$ and $QR$. For $PS$, passing through $(a, 0)$ and $(0, c)$, the intercept form is:

$$ \frac{x}{a} + \frac{y}{c} = 1 \Rightarrow cx + ay = ac $$

For $QR$, passing through $(-\frac{bc}{a}, 0)$ and $(0, b)$, the intercept form is:

$$ \frac{x}{-\frac{bc}{a}} + \frac{y}{b} = 1 \Rightarrow -\frac{ax}{bc} + \frac{y}{b} = 1 \Rightarrow -ax + cy = bc $$

Now, we have a system of two equations with a common parameter $c$. Our mission is to eliminate $c$. From the first equation, $cx + ay = ac$, we can isolate $c$:

$$ c(x - a) = -ay \Rightarrow c = \frac{ay}{a - x} $$

This is the key. We have expressed the 'ghost' parameter $c$ in terms of our coordinates $x$ and $y$.

### The Revelation

Now, we substitute this expression for $c$ into our second equation: $-ax + cy = bc$. Substituting $c = \frac{ay}{a - x}$, we get:

$$ -ax + \left( \frac{ay}{a - x} \right)y = b\left( \frac{ay}{a - x} \right) $$

To clear the denominator, we multiply the entire equation by $(a - x)$. This yields:

$$ -ax(a - x) + ay^2 = aby $$

Expanding this, we get $-a^2x + ax^2 + ay^2 = aby$. Since $a$ is an intercept and not zero, we can divide the entire equation by $a$:

$$ -ax + x^2 + y^2 = by $$

Rearranging this into the standard form, we get:

$$ \mathbf{x^2 + y^2 - ax - by = 0} $$

This, my friend, is the equation of a circle. It passes through the origin because if we set $x=0$ and $y=0$, the equation holds true. We have successfully traced the path of $M$ and found it to be a circle.

Visualized Solution

Setting up the Coordinate System

Defining Line $L_{1}$

Equation and Slope of $L_{1}$

Introducing Perpendicular Line $L_{2}$

Equation of Line $L_{2}$

Coordinates of $R$ and $S$

Drawing Lines $P S$ and $QR$

Equation of Line $P S$

Equation of Line $QR$

Eliminating the Parameter $c$

Substituting $c$ into $QR$

Simplifying the Locus Equation

Final Locus and Conclusion

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Setup

The Geometry of Intersection

The Algebraic Exorcism

The Revelation

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Visualized Solution

Setting up the Coordinate System

Defining Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L1​

Equation and Slope of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L1​

Introducing Perpendicular Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L2​

Equation of Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L2​

Coordinates of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S

Drawing Lines .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PS and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } QR

Equation of Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PS

Equation of Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } QR

Eliminating the Parameter .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c into .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } QR