Let A(α, -2), B(α, 6) and C(α/4, -2) be vertices of a ΔABC. If (5, α/4) is the circumcentre of ΔABC,

Q: How do we know for sure that $\angle A$ is $90^\circ$?

Great question! Look at the segments. $AB$ is a vertical line because both points have the same x-coordinate, $\alpha$. $AC$ is a horizontal line because both points have the same y-coordinate, $-2$. A vertical line and a horizontal line are always perpendicular, meaning they meet at a $90^\circ$ angle at vertex $A$.

Q: Why is the circumcentre the midpoint of the hypotenuse?

This is a classic property of right-angled triangles. The circumcircle of a right triangle has the hypotenuse as its diameter. Since the diameter passes through the center of the circle (the circumcentre), the circumcentre must be the midpoint of that diameter—which is the hypotenuse itself!

Q: Is there a faster way to find the inradius of a right triangle?

Yes! For a right-angled triangle with legs $a$ and $b$ and hypotenuse $c$, the inradius $r$ is given by the formula $r = \frac{a + b - c}{2}$. You can use this to quickly verify your result after calculating the side lengths.

Question

Let $A(\alpha, -2), B(\alpha, 6)$ and $C(\frac{\alpha}{4}, -2)$ be vertices of a $\Delta ABC$. If $(5, \frac{\alpha}{4})$ is the circumcentre of $\Delta ABC$, then which of the following is NOT correct about $\Delta ABC$:

Options:

(A) area is $24$

(B) perimeter is $25$

(C) circumradius is $5$

(D) inradius is $2$

Sigma Percentile · Accepted Answer

### The Geometry of Truth: Unlocking the Secrets of $\Delta ABC$

Welcome, fellow traveler on the path to JEE mastery! Today, we are not just solving a problem; we are peeling back the layers of a geometric mystery.

Coordinate geometry is often seen as a dry collection of formulas, but when you look closer, it is a beautiful, logical dance of points and lines. Let us dive into this problem and see what secrets $\Delta ABC$ holds.

### Phase 1: The Visual Landscape

We are given three vertices: $A(\alpha, -2)$, $B(\alpha, 6)$, and $C(\frac{\alpha}{4}, -2)$. Before we rush into any complex calculations, let us pause and visualize.

Look at $A$ and $B$. They share the same x-coordinate, $\alpha$. This means the segment $AB$ is a perfectly vertical line.

Now look at $A$ and $C$. They share the same y-coordinate, $-2$. This means the segment $AC$ is a perfectly horizontal line.

What happens when a vertical line meets a horizontal line? They intersect at a perfect $90^\circ$ angle.

Just like that, we have identified that $\Delta ABC$ is a right-angled triangle, with the right angle sitting proudly at vertex $A$. This simple observation is our key to the entire problem.

### Phase 2: The Circumcentre's Secret

Now, the problem introduces the circumcentre, located at $(5, \frac{\alpha}{4})$. In the world of geometry, the circumcentre is the center of the circle that passes through all three vertices.

For a general triangle, finding this point can be a nightmare of perpendicular bisectors. But we have a right-angled triangle!

There is a beautiful theorem here: in a right-angled triangle, the circumcentre is always the midpoint of the hypotenuse. Why? Because the hypotenuse acts as the diameter of the circumcircle.

Since the diameter must pass through the center, the center must be the midpoint. Our hypotenuse is the side opposite to the right angle, which is $BC$. So, the circumcentre is simply the midpoint of $BC$.

### Phase 3: The Algebraic Dance

Now that we know the circumcentre is the midpoint of $BC$, let us find its coordinates. Using the midpoint formula, $M = (\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})$, we calculate the midpoint of $B(\alpha, 6)$ and $C(\frac{\alpha}{4}, -2)$:

$$M = \left( \frac{\alpha + \frac{\alpha}{4}}{2}, \frac{6 + (-2)}{2} ight) = \left( \frac{5\alpha}{8}, 2 ight)$$

We are given the circumcentre as $(5, \frac{\alpha}{4})$. Since these two points are the same, we equate their coordinates:

$$\frac{5\alpha}{8} = 5 \implies \alpha = 8$$

$$\frac{\alpha}{4} = 2 \implies \alpha = 8$$

Everything aligns perfectly! We have found $\alpha = 8$.

### Phase 4: The Final Verification

With $\alpha = 8$, our vertices are $A(8, -2)$, $B(8, 6)$, and $C(2, -2)$. Let us calculate the side lengths:

- $AB = |6 - (-2)| = 8$ units 
- $AC = |8 - 2| = 6$ units 
- $BC = \sqrt{8^2 + 6^2} = 10$ units

Now, let us check the options:

1. **Area:** 
$$\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 6 	imes 8 = 24 	ext{ (Correct)}$$

2. **Perimeter:** 
$$8 + 6 + 10 = 24 	ext{ (Wait, the option says } 25 	ext{. This is incorrect!)}$$

3. **Circumradius:** 
$$	ext{Half of the hypotenuse} = \frac{10}{2} = 5 	ext{ (Correct)}$$

4. **Inradius:** 
$$\frac{	ext{Area}}{	ext{Semi-perimeter}} = \frac{24}{12} = 2 	ext{ (Correct)}$$

### Conclusion

The question asked us to find the statement that is NOT correct. We found that the perimeter is $24$, not $25$.

Therefore, **Option (B)** is the one we are looking for. Remember, in JEE, it is not just about the calculation; it is about the conceptual clarity that guides your hand. Keep practicing, keep visualizing, and keep falling in love with the logic!

Animated Solution for Mathematics - Straight Lines: Let $A (α, - 2), B (α, 6)$ and $C (\frac{α}{4}, - 2)$ be vertices of a $Δ A B C$ . If $(5, \frac{α}{4})$ is the circumcentre of $Δ A B C$ , then which of the following is NOT correct about $Δ A B C$ :

Select Answer:

Visualized Solution

Analyzing the Vertices

Identifying the Triangle Type

Circumcentre of a Right Triangle

Calculating the Midpoint

Solving for $α$

Finding Side Lengths

Checking Option (A): Area

Checking Option (B): Perimeter

Checking Option (C): Circumradius

Checking Option (D): Inradius

Final Conclusion

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

The Geometry of Truth

Phase 1

Phase 2

Phase 3

Phase 4

Conclusion

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