If the points P and Q are respectively the circumcenter and the orthocentre of a ΔABC, then PA + PB

Q: Why do we set the circumcenter $P$ as the origin? Couldn't we pick any point?

You absolutely could, but in vector geometry, the choice of origin is like choosing the right vantage point on a mountain. By setting $P$ as the origin, the position vector $\vec{P}$ becomes $\vec{0}$, which simplifies the section formula drastically. It turns a complex algebraic mess into a clean, elegant result.

Q: Wait, is the Euler line property valid for all triangles?

Yes! It is a fundamental property of Euclidean geometry. For any non-equilateral triangle, the orthocenter, centroid, and circumcenter are distinct and collinear. If the triangle is equilateral, these points coincide, and the vector sum is still zero, which is consistent with the formula.

Q: Why is the centroid $G$ defined as the average of the vertices?

The centroid is the geometric center of mass of the triangle. If you place equal masses at each vertex, the center of mass is the average of their positions. Mathematically, this is $\vec{G} = \frac{\vec{a} + \vec{b} + \vec{c}}{3}$. It is a fundamental definition that links the vertices to the interior of the triangle.

Question

If the points $P$ and $Q$ are respectively the circumcenter and the orthocentre of a $\Delta ABC$, then $\vec{PA} + \vec{PB} + \vec{PC}$ is equal to _______

Options:

(A) $2\vec{QP}$

(B) $2\vec{PQ}$

(C) $\vec{PQ}$

(D) $\vec{QP}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

In vector geometry, the choice of origin is your most powerful tool. It is the difference between a page full of messy algebra and a single, elegant line of proof.

We are given the circumcenter $P$. Let us be bold and set $P$ as our origin. By doing this, the position vector of $P$ becomes the zero vector, $\vec{P} = \vec{0}$.

Suddenly, the vectors $\vec{PA}$, $\vec{PB}$, and $\vec{PC}$ are simply the position vectors of the vertices $A$, $B$, and $C$ relative to $P$. Let us call them $\vec{a}$, $\vec{b}$, and $\vec{c}$. Our target expression, $\vec{PA} + \vec{PB} + \vec{PC}$, transforms into the simple sum $\vec{a} + \vec{b} + \vec{c}$.

### The Centroid Bridge

Now, how do we relate these vertices to the orthocenter $Q$? We need a bridge. That bridge is the centroid, $G$.

You likely remember that the centroid is the average of the vertices, defined as:

$$\vec{G} = \frac{\vec{a} + \vec{b} + \vec{c}}{3}$$

If we rearrange this, we find a beautiful identity: $\vec{a} + \vec{b} + \vec{c} = 3\vec{G}$. This is our first major breakthrough. We have successfully linked our target sum to the centroid.

### The Euler Line Revelation

This is where the magic happens. In any triangle, the orthocenter $Q$, the centroid $G$, and the circumcenter $P$ are not scattered randomly; they lie on a single, beautiful line called the **Euler Line**.

They follow a strict ratio: the centroid $G$ divides the segment $PQ$ in a $2:1$ ratio. Mathematically, using the section formula, the position vector of $G$ is given by:

$$\vec{G} = \frac{2\vec{P} + 1\vec{Q}}{2 + 1}$$

Since we brilliantly set $P$ as the origin, $\vec{P} = \vec{0}$. The equation simplifies instantly to $\vec{G} = \frac{\vec{Q}}{3}$. Multiplying by $3$, we get $\vec{Q} = 3\vec{G}$.

### Final Synthesis

Look at what we have achieved! We established that $\vec{PA} + \vec{PB} + \vec{PC} = 3\vec{G}$. We also established that $\vec{Q} = 3\vec{G}$.

By the transitive property of equality, it must be true that:

$$\vec{PA} + \vec{PB} + \vec{PC} = \vec{Q}$$

Since $P$ is the origin, the position vector $\vec{Q}$ is simply the vector $\vec{PQ}$. And there it is—the elegance of geometry revealed. The sum of the vectors from the circumcenter to the vertices is exactly the vector from the circumcenter to the orthocenter, or **$\vec{PQ}$**.

Animated Solution for Mathematics - Straight Lines: If the points $P$ and $Q$ are respectively the circumcenter and the orthocentre of a $Δ A B C$ , then $P A + P B + P C$ is equal to _______

Select Answer:

Visualized Solution

Visualizing the Triangle and Circumcenter

Locating the Orthocenter

Setting the Origin

Defining Position Vectors

The Target Expression

Introducing the Centroid

Relating Sum to Centroid

The Euler Line Property

Section Formula on Euler Line

Applying the Section Formula

Substituting the Origin

Expressing Orthocenter via Centroid

Equating the Expressions

Final Vector Representation

The Sigma Insight: Centroid, Incenter, Orthocenter, and Circumcenter

Analyzing the Setup

The Centroid Bridge

The Euler Line Revelation

Final Synthesis

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Animated Solution for Mathematics - Straight Lines: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the points P and Q are respectively the circumcenter and the orthocentre of a ΔABC, then PA+PB+PC is equal to _______