Let A(1, 0), B(6, 2) and C(3/2, 6) be the vertices of a triangle ABC. If P is a point inside the tri

Q: Wait, why is $P$ the centroid? Could it be the incenter or the orthocenter?

That is a fantastic question! In geometry, the centroid is the unique point where the medians intersect, and it is the only point that divides the triangle into three triangles of equal area. If you were to choose the incenter or orthocenter, the areas of the resulting triangles would depend on the side lengths, not just the vertices. The equal area property is a defining characteristic of the centroid.

Q: Is there a faster way to find the distance without calculating the centroid coordinates?

While you might be tempted to look for a shortcut, calculating the centroid is actually the most direct and robust path here. The coordinates of the vertices are simple enough that finding the centroid takes only a few seconds. Trust the process—the algebra is designed to simplify beautifully!

Q: What if the point $Q$ was outside the triangle? Does the distance formula still work?

Absolutely! The distance formula $d = \sqrt{(x_P - x_Q)^2 + (y_P - y_Q)^2}$ is a universal tool in coordinate geometry. It doesn't care whether the points are inside, outside, or on the boundary of the triangle. As long as you have the coordinates, the formula will give you the correct Euclidean distance.

Question

Let $A(1, 0)$, $B(6, 2)$ and $C\left(\frac{3}{2}, 6\right)$ be the vertices of a triangle $ABC$. If $P$ is a point inside the triangle $ABC$ such that the triangles $APC$, $APB$ and $BPC$ have equal areas, then the length of the line segment $PQ$, where $Q$ is the point $\left(-\frac{7}{6}, -\frac{1}{3}\right)$, is.....

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the path to JEE mastery! Today, we are going to explore a beautiful problem that bridges the gap between pure geometry and coordinate algebra.

Imagine you are standing on a coordinate plane, looking at a triangle $ABC$ with vertices $A(1, 0)$, $B(6, 2)$, and $C\left(\frac{3}{2}, 6\right)$. This is our geometric playground where we uncover the hidden symmetry of the triangle.

### The Mystery of Point $P$

We are introduced to a point $P$ inside this triangle. The problem provides a crucial clue: the triangles $APC$, $APB$, and $BPC$ all have equal areas.

If you connect $P$ to the vertices, you are partitioning the triangle into three perfectly balanced regions. In the world of geometry, there is only one point that possesses this property: the **centroid**.

The point where the medians of the triangle intersect is the very same point that divides the triangle into three equal areas. Understanding this connection is the key that unlocks the entire problem.

### The Centroid Calculation

Now that we know $P$ is the centroid, finding its coordinates is straightforward. We use the centroid formula, which is the average of the vertices' coordinates.

For the $x$-coordinate, we calculate:

$$x_P = \frac{1 + 6 + \frac{3}{2}}{3} = \frac{\frac{17}{2}}{3} = \frac{17}{6}$$

Similarly, for the $y$-coordinate, we take the average of the $y$-values:

$$y_P = \frac{0 + 2 + 6}{3} = \frac{8}{3}$$

So, our centroid $P$ is located at $\left(\frac{17}{6}, \frac{8}{3}\right)$.

### The Final Stretch

Our goal is to find the length of the segment $PQ$, where $Q$ is the point $\left(-\frac{7}{6}, -\frac{1}{3}\right)$. We reach for our trusty distance formula:

$$PQ = \sqrt{(x_P - x_Q)^2 + (y_P - y_Q)^2}$$

Let's substitute our values:

$$PQ = \sqrt{\left(\frac{17}{6} - \left(-\frac{7}{6}\right)\right)^2 + \left(\frac{8}{3} - \left(-\frac{1}{3}\right)\right)^2}$$

The $x$-difference becomes $\frac{17}{6} + \frac{7}{6} = \frac{24}{6} = 4$. The $y$-difference becomes $\frac{8}{3} + \frac{1}{3} = \frac{9}{3} = 3$.

The equation simplifies to:

$$PQ = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25}$$

The distance is exactly **$5$** units. You have successfully navigated the geometry, applied the centroid property, and conquered the algebra.

Enter Numerical Value:

Visualized Solution

The Triangle $A B C$

Property of Point $P$

$P$ is the Centroid

Centroid $x$ -coordinate Setup

Calculating $x_{P}$

Centroid $y$ -coordinate Setup

Calculating $y_{P}$

Point $Q$ and the Goal

The Distance Formula

Substituting $P$ and $Q$

Simplifying $x$ -difference

Simplifying $y$ -difference

Squaring the Terms

Final Calculation

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

Analyzing the Setup

The Mystery of Point $P$

The Centroid Calculation

The Final Stretch

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Enter Numerical Value:

Visualized Solution

The Triangle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ABC

Property of Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P is the Centroid

Centroid .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x-coordinate Setup

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

Centroid .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y-coordinate Setup

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

Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q and the Goal