The x-coordinate of the incentre of the triangle that has the coordinates of mid points of its sides

Q: Wait, why can't we just use the centroid formula for the incenter?

That is a classic trap! The centroid is the intersection of medians, while the incenter is the intersection of angle bisectors. They are only the same in an equilateral triangle. For any other triangle, they are distinct points. You must use the specific incenter formula involving side lengths.

Q: Is there a faster way to find the vertices without solving all those equations?

While you can sometimes use geometric properties, solving the system of equations is the most robust and reliable method for JEE Advanced. It ensures you don't miss any subtle coordinate shifts. It's a standard technique that builds your algebraic fluency.

Q: Why did the denominator simplify so nicely in the final step?

That is the beauty of this specific problem! Because the vertices form an isosceles right-angled triangle, the side lengths have a specific ratio involving $\sqrt{2}$. When you plug them into the incenter formula, the terms naturally align to allow for simplification. Always look for these patterns; they are often a sign that you are on the right track.

Question

The x-coordinate of the incentre of the triangle that has the coordinates of mid points of its sides as $(0, 1), (1, 1)$ and $(1, 0)$ is

Options:

(A) $2 + \sqrt{2}$

(B) $2 - \sqrt{2}$

(C) $1 + \sqrt{2}$

(D) $1 - \sqrt{2}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a coordinate plane, looking at three points: $M_1(0, 1)$, $M_2(1, 1)$, and $M_3(1, 0)$. These are the midpoints of the sides of a triangle that we cannot yet see.

Our mission is to find the x-coordinate of the incenter of this invisible, larger triangle. This is a classic JEE Advanced challenge that tests your ability to bridge the gap between given data and the hidden structure of the problem.

### Unveiling the Vertices

We start by assuming the vertices of our mystery triangle are $A(x_1, y_1)$, $B(x_2, y_2)$, and $C(x_3, y_3)$. The midpoint formula provides the following system of equations for the x-coordinates:

$$\frac{x_1 + x_2}{2} = 0, \quad \frac{x_2 + x_3}{2} = 1, \quad \frac{x_3 + x_1}{2} = 1$$

This simplifies to $x_1 + x_2 = 0$, $x_2 + x_3 = 2$, and $x_3 + x_1 = 2$. Adding these three equations yields:

$$2(x_1 + x_2 + x_3) = 4 \implies x_1 + x_2 + x_3 = 2$$

By subtracting each original equation from this sum, we find $x_1 = 0$, $x_2 = 0$, and $x_3 = 2$. Repeating this logic for the y-coordinates yields $y_1 = 0$, $y_2 = 2$, and $y_3 = 0$.

The mystery clears: our vertices are $A(0, 0)$, $B(0, 2)$, and $C(2, 0)$. We have uncovered a right-angled triangle sitting at the origin.

### Measuring the Sides

Now that we know the vertices, we calculate the side lengths using the distance formula:

$$a = BC = \sqrt{(2-0)^2 + (0-2)^2} = \sqrt{8} = 2\sqrt{2}$$

$$b = AC = \sqrt{(2-0)^2 + (0-0)^2} = 2$$

$$c = AB = \sqrt{(0-0)^2 + (2-0)^2} = 2$$

We have an isosceles right-angled triangle with sides $2\sqrt{2}$, $2$, and $2$.

### The Incenter Calculation

The incenter $I(x, y)$ has an x-coordinate given by the formula:

$$x = \frac{ax_1 + bx_2 + cx_3}{a + b + c}$$

Substituting our known values into the equation:

$$x = \frac{(2\sqrt{2})(0) + (2)(0) + (2)(2)}{2\sqrt{2} + 2 + 2} = \frac{4}{4 + 2\sqrt{2}}$$

Dividing both the numerator and denominator by $2$, we obtain:

$$x = \frac{2}{2 + \sqrt{2}}$$

To finalize, we rationalize the denominator by multiplying by the conjugate $(2 - \sqrt{2})$:

$$x = \frac{2(2 - \sqrt{2})}{(2 + \sqrt{2})(2 - \sqrt{2})} = \frac{2(2 - \sqrt{2})}{4 - 2} = \frac{2(2 - \sqrt{2})}{2} = 2 - \sqrt{2}$$

The x-coordinate of the incenter is **$2 - \sqrt{2}$**.

Animated Solution for Mathematics - Straight Lines: The x-coordinate of the incentre of the triangle that has the coordinates of mid points of its sides as $(0, 1), (1, 1)$ and $(1, 0)$ is

Select Answer:

Visualized Solution

Plotting the Midpoints

Connecting Midpoints to Vertices

Setting up the Equations

Solving for $x_{1}, x_{2}, x_{3}$

Solving for $y_{1}, y_{2}, y_{3}$

Calculating Side Lengths

The Incenter Formula

Substituting the Values

Simplifying the Fraction

Rationalizing the Denominator

Final Calculation

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

Analyzing the Setup

Unveiling the Vertices

Measuring the Sides

The Incenter Calculation

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