Let the circumcentre of a triangle with vertices A(a, 3), B(b, 5) and C(a, b), ab 0 be P(1, 1). If

Q: Why do we square the distances? Can't we just use the distance formula with square roots?

You absolutely could, but it is a classic trap! Working with square roots often leads to messy algebraic expressions and increases the likelihood of calculation errors. By squaring both sides ($PA^2 = PB^2 = PC^2$), we linearize the equations, making the algebra much cleaner and more manageable.

Q: I found two possible values for $a$ and $b$. How do I know which ones to choose?

This is a brilliant question. In coordinate geometry, always look for constraints. The problem explicitly states $ab > 0$. Furthermore, you must check for geometric validity: if a specific pair of $a$ and $b$ causes two vertices to coincide (e.g., $A$ and $C$ becoming the same point), the figure is no longer a triangle. Always test your solutions against the physical reality of the problem!

Q: Is there a faster way to find the circumcenter than solving for $a$ and $b$?

While there are many properties of circumcenters, in this problem, the vertices themselves are defined by the variables $a$ and $b$. Since the circumcenter $P$ is already given, solving for $a$ and $b$ is the most direct and reliable path. Don't let the complexity of the geometry intimidate you; sometimes, the most elegant solution is simply to let the algebra reveal the structure.

Question

Let the circumcentre of a triangle with vertices $A(a, 3), B(b, 5)$ and $C(a, b), ab > 0$ be $P(1, 1)$. If the line $AP$ intersects the line $BC$ at the point $Q(k_1, k_2)$, then $k_1 + k_2$ is equal to :

Options:

(A) $2$

(B) $\frac{4}{7}$

(C) $\frac{2}{7}$

(D) $4$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on the Cartesian plane, looking at a triangle defined by vertices $A(a, 3)$, $B(b, 5)$, and $C(a, b)$. We are given a fixed anchor: the circumcenter $P(1, 1)$.

The circumcenter is the heart of the triangle, the unique point that sits at the center of the circle passing through all three vertices. Because it is the center, the distance from $P$ to $A$, $P$ to $B$, and $P$ to $C$ must be identical.

This is our golden key: $PA = PB = PC$. To make our lives easier, we work with the squares of these distances, $PA^2 = PB^2 = PC^2$, avoiding the cumbersome square roots that often lead to errors.

### The Algebraic Dance

Let us translate this geometric truth into the language of algebra. Using the distance formula $d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$, we write out our expressions:

$$PA^2 = (a - 1)^2 + (3 - 1)^2 = (a - 1)^2 + 4$$

$$PB^2 = (b - 1)^2 + (5 - 1)^2 = (b - 1)^2 + 16$$

$$PC^2 = (a - 1)^2 + (b - 1)^2$$

Now, watch the magic happen. When we equate $PA^2 = PC^2$, the $(a - 1)^2$ term appears on both sides and vanishes, leaving us with $4 = (b - 1)^2$. This implies $b - 1 = \pm 2$, giving us $b = 3$ or $b = -1$.

Similarly, equating $PB^2 = PC^2$ causes the $(b - 1)^2$ term to cancel, leaving $16 = (a - 1)^2$, which means $a - 1 = \pm 4$, so $a = 5$ or $a = -3$. We have our candidates, but we must be careful.

### The Geometric Trap

The problem gives us a vital constraint: $ab > 0$. Both pairs $(5, 3)$ and $(-3, -1)$ satisfy this.

However, we must check for geometric integrity. If we choose $a = 5$ and $b = 3$, the coordinates of $A$ become $(5, 3)$ and $C$ become $(5, 3)$. A triangle cannot exist if two of its vertices are the same point!

Thus, we reject this pair. We are left with the only valid solution: $a = -3$ and $b = -1$. Our triangle is now fully defined: $A(-3, 3)$, $B(-1, 5)$, and $C(-3, -1)$.

### The Final Intersection

With the vertices locked in, the rest of the journey is a beautiful exercise in linear equations. We need the intersection of line $AP$ and line $BC$.

For line $AP$, passing through $A(-3, 3)$ and $P(1, 1)$, the slope is:

$$m_{AP} = \frac{1 - 3}{1 - (-3)} = -\frac{1}{2}$$

Using the point-slope form, we derive the equation $x + 2y = 3$. For line $BC$, passing through $B(-1, 5)$ and $C(-3, -1)$, the slope is:

$$m_{BC} = \frac{-1 - 5}{-3 - (-1)} = 3$$

This gives us the equation $y = 3x + 8$. Solving this system by substituting $y$ into the first equation, we find:

$$x + 2(3x + 8) = 3 \Rightarrow 7x = -13 \Rightarrow x = -\frac{13}{7}$$

Substituting this back, we get $y = \frac{17}{7}$. The sum $k_1 + k_2 = -\frac{13}{7} + \frac{17}{7} = \frac{4}{7}$.

It is a result that emerges with perfect clarity from the chaos of the initial variables. The final answer is **$\frac{4}{7}$**.

Animated Solution for Mathematics - Straight Lines: Let the circumcentre of a triangle with vertices $A (a, 3), B (b, 5)$ and $C (a, b), ab > 0$ be $P (1, 1)$ . If the line $A P$ intersects the line $B C$ at the point $Q (k_{1}, k_{2})$ , then $k_{1} + k_{2}$ is equal to :

Select Answer:

Visualized Solution

Visualizing the Given Data

The Circumcenter Property

Applying the Distance Formula

Solving for $b$

Solving for $a$

Filtering the Valid Coordinates

The True Triangle Vertices

Finding the Slope of Line $A P$

Equation of Line $A P$

Finding the Slope of Line $B C$

Equation of Line $B C$

Setting up the Intersection

Calculating Coordinates of $Q$

Final Answer: $k_{1} + k_{2}$

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

Analyzing the Setup

The Algebraic Dance

The Geometric Trap

The Final Intersection

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Select Answer:

Visualized Solution

Visualizing the Given Data

The Circumcenter Property

Applying the Distance Formula

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

Filtering the Valid Coordinates

The True Triangle Vertices

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

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

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

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