Let the algebraic sum of the perpendicular distances from the points (2, 0), (0, 2) and (1, 1) to a

Q: Wait, why do we reject the absolute value sign in the distance formula?

Great question! The standard distance formula $d = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}$ uses the absolute value to ensure distance is always positive. However, the problem specifically asks for the 'algebraic sum' of the distances. This implies we are treating the distance as a signed quantity, $d = \frac{ax_0 + by_0 + c}{\sqrt{a^2 + b^2}}$. This is a common technique in coordinate geometry to handle problems involving lines passing through fixed points.

Q: Can we just cancel the $\sqrt{a^2 + b^2}$ term?

Yes, absolutely! Since the sum of the distances is zero, we have $\frac{1}{\sqrt{a^2 + b^2}} \times (\text{sum of numerators}) = 0$. As long as the line exists, $a$ and $b$ cannot both be zero, so $\sqrt{a^2 + b^2}$ is strictly positive. Therefore, we can safely multiply both sides by this term to eliminate it, leaving us with a much simpler linear equation.

Q: How does $a + b + c = 0$ prove the line passes through $(1, 1)$?

Think of the equation $ax + by + c = 0$. If we substitute the point $(1, 1)$ into this equation, we get $a(1) + b(1) + c = 0$, which simplifies to $a + b + c = 0$. Since our derivation showed that $a + b + c = 0$ must hold for the line to satisfy the given condition, it means that for any values of $a, b,$ and $c$ that satisfy this condition, the point $(1, 1)$ will always satisfy the equation of the line. Thus, the line must pass through $(1, 1)$.

Question

Let the algebraic sum of the perpendicular distances from the points $(2, 0), (0, 2)$ and $(1, 1)$ to a variable straight line be zero; then the line passes through a fixed point whose coordinates are .........

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a vast, empty Cartesian plane. Before you lie three distinct markers: $P_1(2, 0)$ resting on the x-axis, $P_2(0, 2)$ perched on the y-axis, and $P_3(1, 1)$ sitting comfortably in the first quadrant.

These points are the anchors of our problem. We are tasked with finding a 'fixed point' through which a variable line must pass, given that the algebraic sum of the perpendicular distances from these three points to the line is zero.

### The Concept of Algebraic Distance

First, we must clarify what 'algebraic sum' means. In geometry, the distance from a point $(x_0, y_0)$ to a line $ax + by + c = 0$ is typically defined as $d = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}$.

However, the term 'algebraic' instructs us to drop the absolute value bars. We are looking at the signed distance:

$$d = \frac{ax_0 + by_0 + c}{\sqrt{a^2 + b^2}}$$

This means if a point lies on one side of the line, its contribution to the sum is positive; if it lies on the other, it is negative. The condition that the sum is zero implies a perfect balance where the positive and negative distances cancel each other out.

### The Mathematical Dance

Let us write down the condition for our three points. For $P_1(2, 0)$, $P_2(0, 2)$, and $P_3(1, 1)$, the distances are:

$$d_1 = \frac{2a + c}{\sqrt{a^2 + b^2}}, \quad d_2 = \frac{2b + c}{\sqrt{a^2 + b^2}}, \quad d_3 = \frac{a + b + c}{\sqrt{a^2 + b^2}}$$

The problem states that $d_1 + d_2 + d_3 = 0$. When we sum these, we obtain:

$$\frac{2a + c}{\sqrt{a^2 + b^2}} + \frac{2b + c}{\sqrt{a^2 + b^2}} + \frac{a + b + c}{\sqrt{a^2 + b^2}} = 0$$

Since the denominator $\sqrt{a^2 + b^2}$ is common to all terms, we can multiply the entire equation by this value to remove it. We are left with the sum of the numerators:

$$(2a + c) + (2b + c) + (a + b + c) = 0$$

### The Revelation

Now, let us group the terms. We have $a$ terms ($2a + a = 3a$), $b$ terms ($2b + b = 3b$), and $c$ terms ($c + c + c = 3c$). The equation simplifies to:

$$3a + 3b + 3c = 0$$

Dividing by $3$, we arrive at the elegant relation:

$$a + b + c = 0$$

This is the secret key. Our variable line is defined by $ax + by + c = 0$. If we rewrite our condition as $a(1) + b(1) + c = 0$, we see that this is exactly the result of plugging the point $(1, 1)$ into the line equation.

Regardless of the values of $a, b,$ and $c$, as long as they satisfy $a + b + c = 0$, the line will always pass through the point **$(1, 1)$**. You have successfully identified the invariant fixed point that remains unmoved amidst the change.

Animated Solution for Mathematics - Straight Lines: Let the algebraic sum of the perpendicular distances from the points $(2, 0), (0, 2)$ and $(1, 1)$ to a variable straight line be zero; then the line passes through a fixed point whose coordinates are .........

Visualized Solution

Plotting the Given Points

The Variable Straight Line

Algebraic Perpendicular Distance

Setting Up the Sum Condition

Substituting the Coordinates

Eliminating the Denominator

Grouping Like Terms

Simplifying the Relation

Comparing with the Line Equation

Identifying the Fixed Point

The Sigma Insight: Family of Lines

Analyzing the Setup

The Concept of Algebraic Distance

The Mathematical Dance

The Revelation

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