Let C(α, β) be the circumcentre of the triangle formed by the lines 4x + 3y = 69, 4y - 3x = 17, and

Q: Wait, why is the circumcentre always the midpoint of the hypotenuse in a right-angled triangle?

This is a beautiful consequence of Thales' Theorem! In a right-angled triangle, the hypotenuse acts as the diameter of the circumcircle. Since the circumcentre is the center of this circle, it must lie at the midpoint of the diameter, which is the hypotenuse. It is a classic JEE shortcut that saves you from solving for the intersection of perpendicular bisectors.

Q: How do I know which side is the hypotenuse?

The hypotenuse is always the side opposite the right angle. Since we found that $L_1$ and $L_2$ are perpendicular, their intersection is the vertex where the right angle is located. Therefore, the third line, $L_3$, which does not pass through this intersection, must be the hypotenuse.

Q: Can I just use the general formula for the circumcentre of a triangle?

You could, but it would be a massive waste of time! The general method involves finding the intersection of two perpendicular bisectors, which requires finding the midpoints of two sides, calculating the slopes of those sides, finding the negative reciprocal slopes, and then solving a system of linear equations. By recognizing the right-angled triangle property, you reduce a 10-minute calculation to a 2-minute one. Always look for the geometry first!

Question

Let $C(\alpha, \beta)$ be the circumcentre of the triangle formed by the lines $4x + 3y = 69$, $4y - 3x = 17$, and $x + 7y = 61$. Then $(\alpha - \beta)^2 + \alpha + \beta$ is equal to

Options:

(A) 18

(B) 17

(C) 15

(D) 16

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a triangle defined by three lines: 
$L_1: 4x + 3y = 69$, 
$L_2: 4y - 3x = 17$, and 
$L_3: x + 7y = 61$.

At first glance, this looks like a standard, perhaps tedious, coordinate geometry problem. You might be tempted to immediately start solving for the vertices by intersecting these lines.

But wait! A true JEE aspirant knows that before you start calculating, you must look for the pattern. Let us analyze the slopes.

For $L_1$, we rewrite it as $y = -\frac{4}{3}x + 23$, giving us a slope $m_1 = -\frac{4}{3}$. 
For $L_2$, we rewrite it as $y = \frac{3}{4}x + \frac{17}{4}$, giving us a slope $m_2 = \frac{3}{4}$.

Now, look at the product: 
$$m_1 \cdot m_2 = \left(-\frac{4}{3}\right) \cdot \left(\frac{3}{4}\right) = -1$$

This is the "**Aha!**" moment. The product of the slopes is $-1$, which means $L_1$ and $L_2$ are perpendicular. Our triangle is a **right-angled triangle**!

### The Elegant Shortcut

Now that we know the triangle is right-angled, we do not need to perform the grueling task of finding the intersection of perpendicular bisectors.

We invoke the elegant property of the circumcentre in a right-angled triangle: it lies exactly at the **midpoint of the hypotenuse**. The hypotenuse is the side opposite the right angle, which is the side formed by $L_3$.

Our task is now reduced to finding the two endpoints of the hypotenuse, which are the intersections of $L_3$ with $L_1$ and $L_2$.

### The Final Calculation

First, let us find vertex $B$, the intersection of $L_2$ and $L_3$. Solving $-3x + 4y = 17$ and $x + 7y = 61$ simultaneously, we find $B(5, 8)$.

Next, we find vertex $C$, the intersection of $L_1$ and $L_3$. Solving $4x + 3y = 69$ and $x + 7y = 61$ simultaneously, we find $C(12, 7)$.

The circumcentre $M(\alpha, \beta)$ is simply the midpoint of $BC$: 
$$\alpha = \frac{5 + 12}{2} = \frac{17}{2}, \quad \beta = \frac{8 + 7}{2} = \frac{15}{2}$$

We have our coordinates: $(\frac{17}{2}, \frac{15}{2})$.

Finally, we evaluate the expression $(\alpha - \beta)^2 + \alpha + \beta$. Substituting our values, we get: 
$$\left(\frac{17}{2} - \frac{15}{2}\right)^2 + \frac{17}{2} + \frac{15}{2} = (1)^2 + \frac{32}{2} = 1 + 16 = 17$$

The beauty of this problem lies not in the arithmetic, but in the recognition of the geometric structure. By spotting the perpendicularity early, we bypassed the complexity and arrived at the answer **17** with precision and grace.

Animated Solution for Mathematics - Straight Lines: Let $C (α, β)$ be the circumcentre of the triangle formed by the lines $4 x + 3 y = 69$ , $4 y - 3 x = 17$ , and $x + 7 y = 61$ . Then $(α - β)^{2} + α + β$ is equal to

Select Answer:

Visualized Solution

The Triangle Setup

Analyzing the Slopes

The Perpendicularity Check

Circumcentre of a Right Triangle

Finding Vertex $B$

Finding Vertex $C$

Locating the Circumcentre

Calculating $α$ and $β$

Setting Up the Final Expression

Final Evaluation

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

Analyzing the Setup

The Elegant Shortcut

The Final Calculation

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