Consider a triangle having vertices A(-2, 3), B(1, 9) and C(3, 8). If a line L passing through the c

Q: Why can't I just find the perpendicular bisectors of $AB$ and $BC$ to find the circumcenter?

You absolutely can! However, in competitive exams like JEE, time is your most precious resource. Recognizing that the triangle is right-angled allows you to bypass the tedious algebra of finding two perpendicular bisector equations and solving them simultaneously. Always look for geometric properties first.

Q: How did you know it was a right-angled triangle so quickly?

It's a classic pattern. Whenever you see coordinates, quickly calculate the squared distances of the sides. If the sum of the squares of two sides equals the square of the third, you have a right-angled triangle. Here, $AB^2 + BC^2 = 45 + 5 = 50 = AC^2$ was the giveaway.

Q: What if the line didn't pass through the midpoint of $BC$?

If the line didn't pass through the midpoint, the problem would have provided a different constraint, such as a specific slope or another point. Always read the problem statement carefully to identify the constraints given for the line $L$.

Question

Consider a triangle having vertices $A(-2, 3), B(1, 9)$ and $C(3, 8)$. If a line $L$ passing through the circum-centre of triangle $ABC$, bisects line $BC$, and intersects y-axis at point $(0, \frac{\alpha}{2})$, then the value of real number $\alpha$ is

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on the Cartesian plane, looking at three points: $A(-2, 3)$, $B(1, 9)$, and $C(3, 8)$. At first glance, they are just coordinates, but to a mathematician, they are the vertices of a story waiting to be told.

Our mission is to find the value of $\alpha$. Before we dive into the algebra, we must understand the geometry. The first step in any coordinate geometry problem is to visualize the shape.

Let's calculate the squared lengths of the sides using the distance formula, $d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$:

$$AB^2 = (1 - (-2))^2 + (9 - 3)^2 = 3^2 + 6^2 = 9 + 36 = 45$$

$$BC^2 = (3 - 1)^2 + (8 - 9)^2 = 2^2 + (-1)^2 = 4 + 1 = 5$$

$$AC^2 = (3 - (-2))^2 + (8 - 3)^2 = 5^2 + 5^2 = 25 + 25 = 50$$

Do you see the beauty in these numbers? Since $45 + 5 = 50$, this is the Pythagoras theorem in action: $AB^2 + BC^2 = AC^2$. Our triangle is **right-angled at vertex $B$**. This realization is our first major breakthrough.

### The Geometric Shortcut: Locating the Circumcenter

Now that we know $	riangle ABC$ is a right-angled triangle, we have a powerful property at our disposal. In any right-angled triangle, the circumcenter—the point equidistant from all vertices—lies exactly at the midpoint of the hypotenuse.

Since our hypotenuse is $AC$, the circumcenter $O$ is simply the midpoint of $AC$:

$$O = \left( \frac{-2 + 3}{2}, \frac{3 + 8}{2} ight) = \left( \frac{1}{2}, \frac{11}{2} ight)$$

This is much faster than finding the intersection of two perpendicular bisectors. This is the kind of insight that separates the good from the great in JEE Advanced.

### Constructing the Line $L$

The problem states that line $L$ passes through the circumcenter $O$ and bisects the side $BC$. Bisecting $BC$ means the line must pass through the midpoint of $BC$. Let's call this midpoint $M$:

$$M = \left( \frac{1 + 3}{2}, \frac{9 + 8}{2} ight) = \left( 2, \frac{17}{2} ight)$$

Now, we have two points that define our line $L$: $O\left(\frac{1}{2}, \frac{11}{2}ight)$ and $M\left(2, \frac{17}{2}ight)$. To find the equation of this line, we first calculate its slope $m$:

$$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{\frac{17}{2} - \frac{11}{2}}{2 - \frac{1}{2}} = \frac{\frac{6}{2}}{\frac{3}{2}} = \frac{3}{1.5} = 2$$

With the slope $m=2$ and point $O\left(\frac{1}{2}, \frac{11}{2}ight)$, we use the point-slope form:

$$y - \frac{11}{2} = 2\left(x - \frac{1}{2}ight)$$

Simplifying this, we get $y = 2x - 1 + \frac{11}{2}$, which results in the line equation:

$$y = 2x + \frac{9}{2}$$

### The Final Reveal

The problem states that line $L$ intersects the y-axis at $(0, \frac{\alpha}{2})$. On the y-axis, the x-coordinate is always zero.

Substituting $x=0$ into our equation $y = 2x + \frac{9}{2}$, we find $y = \frac{9}{2}$. By comparing this to the given y-coordinate $\frac{\alpha}{2}$, we set up the equality:

$$\frac{\alpha}{2} = \frac{9}{2}$$

This leads us directly to **$\alpha = 9$**. Through careful geometric analysis and systematic algebraic steps, we have arrived at the solution.

Enter Numerical Value:

Visualized Solution

Visualize the Triangle $A B C$

Analyzing the Triangle's Sides

Calculate Side Lengths Squared

Verify Right-Angled Property

Circumcenter of a Right Triangle

Locate Circumcenter $O$

Line $L$ Bisects $B C$

Find Midpoint $M$ of $B C$

Slope of Line $L$

Calculate Slope $m$

Equation of Line $L$

Find $y$ -intercept and $α$

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

Analyzing the Setup

The Geometric Shortcut

Constructing the Line $L$

The Final Reveal

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Enter Numerical Value:

Visualized Solution

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

Analyzing the Triangle's Sides

Calculate Side Lengths Squared

Verify Right-Angled Property

Circumcenter of a Right Triangle

Locate Circumcenter .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } O

Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L Bisects .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } BC

Find Midpoint .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } BC

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

Calculate Slope .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m

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

Find .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y-intercept and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α