Two vertices of a triangle ABC are A(3, -1) and B(-2, 3), and its orthocentre is P(1, 1). If the coo

Q: Wait, why do we need to find the equations of lines $BC$ and $AC$ separately? Can't we just use the properties of the orthocenter directly?

That is a great question! While the orthocenter has many beautiful properties, in this coordinate geometry problem, we are given the coordinates of the orthocenter $P$ and two vertices $A$ and $B$. The most direct path to finding the third vertex $C$ is to treat it as the intersection of the two lines $BC$ and $AC$. Since we know $BC \perp AP$ and $AC \perp BP$, finding these lines is the most robust and error-proof method.

Q: Why did the $h^2$ and $k^2$ terms cancel out when finding the circumcenter? Is that supposed to happen?

Yes, absolutely! It is a beautiful feature of this problem. When you equate the squared distances, you are essentially setting up a quadratic equation in $h$ and $k$. However, because the quadratic terms ($h^2$ and $k^2$) appear on both sides of the equation with the same coefficients, they subtract to zero. This leaves you with a linear equation, which is much easier to solve. It is the geometry's way of being kind to you!

Q: I'm getting confused with the negative signs in the slope calculations. Any tips?

Coordinate geometry is a game of precision. When calculating slopes like $m_{AP} = \frac{1 - (-1)}{1 - 3}$, always write out the signs explicitly. Use parentheses for negative coordinates. It is a small habit, but it prevents the most common 'silly' mistakes that cost marks in JEE Advanced. Take a deep breath, write it down step-by-step, and the signs will fall into place.

Question

Two vertices of a triangle $ABC$ are $A(3, -1)$ and $B(-2, 3)$, and its orthocentre is $P(1, 1)$. If the coordinates of the point $C$ are $(\alpha, \beta)$ and the centre of the circle circumscribing the triangle $PAB$ is $(h, k)$, then the value of $(\alpha + \beta) + 2(h + k)$ equals

Options:

(A) 5

(B) 81

(C) 15

(D) 51

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are given the vertices $A(3, -1)$ and $B(-2, 3)$ of a triangle, along with its orthocenter $P(1, 1)$. Our objective is to determine the third vertex $C(\alpha, \beta)$ and the circumcenter $(h, k)$ of $	riangle PAB$.

This problem relies on the fundamental property that the altitude from a vertex is perpendicular to the opposite side. We will use this to define the lines containing the sides of the triangle.

### The Lines of Altitude

First, we calculate the slope of the altitude $AP$ using the coordinates $A(3, -1)$ and $P(1, 1)$:

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

Since the side $BC$ is perpendicular to $AP$, its slope $m_{BC}$ must satisfy $m_{BC} \cdot m_{AP} = -1$. Therefore, $m_{BC} = 1$. Using the point-slope form for the line passing through $B(-2, 3)$:

$$y - 3 = 1(x + 2) \Rightarrow y - x = 5 \quad \dots (1)$$

Next, we find the slope of the altitude $BP$ using $B(-2, 3)$ and $P(1, 1)$:

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

Since the side $AC$ is perpendicular to $BP$, its slope $m_{AC}$ is the negative reciprocal, $\frac{3}{2}$. Using the point-slope form for the line passing through $A(3, -1)$:

$$y + 1 = \frac{3}{2}(x - 3) \Rightarrow 2y + 2 = 3x - 9 \Rightarrow 3x - 2y = 11 \quad \dots (2)$$

### The Intersection of Destiny

Vertex $C(\alpha, \beta)$ is the intersection of lines (1) and (2). We substitute $y = x + 5$ into the second equation:

$$3x - 2(x + 5) = 11$$

$$3x - 2x - 10 = 11 \Rightarrow x = 21$$

Substituting $x = 21$ back into the first equation, we find $y = 21 + 5 = 26$. Thus, $C(21, 26)$, which gives $\alpha = 21$ and $\beta = 26$. The sum is **$\alpha + \beta = 47$**.

### The Circumcenter of $	riangle PAB$

The circumcenter $(h, k)$ is equidistant from $P(1, 1)$, $A(3, -1)$, and $B(-2, 3)$. We equate the squared distances $PA^2 = PB^2$:

$$(h - 1)^2 + (k - 1)^2 = (h - 3)^2 + (k + 1)^2$$

Expanding and simplifying, the quadratic terms cancel out:

$$-2h + 1 - 2k + 1 = -6h + 9 + 2k + 1 \Rightarrow 4h - 4k = 8 \Rightarrow h - k = 2 \quad \dots (3)$$

Next, we equate the squared distances $PA^2 = PB^2$ (or $PB^2 = PA^2$):

$$(h - 1)^2 + (k - 1)^2 = (h + 2)^2 + (k - 3)^2$$

Expanding this yields:

$$-2h + 1 - 2k + 1 = 4h + 4 - 6k + 9 \Rightarrow 6h - 4k = -11 \quad \dots (4)$$

Solving the system of equations (3) and (4) by substituting $h = k + 2$ into (4):

$$6(k + 2) - 4k = -11 \Rightarrow 2k + 12 = -11 \Rightarrow k = -\frac{23}{2}$$

Consequently, $h = -\frac{23}{2} + 2 = -\frac{19}{2}$.

### Final Calculation

We are tasked with evaluating the expression $(\alpha + \beta) + 2(h + k)$. Substituting our derived values:

$$(\alpha + \beta) = 47$$

$$2(h + k) = 2\left(-\frac{19}{2} - \frac{23}{2}ight) = 2\left(-\frac{42}{2}ight) = -42$$

Combining these results:

$$47 - 42 = 5$$

The final answer is **5**.

Select Answer:

Visualized Solution

Visualize the Given Points

Slope of $A P$

Equation of side $B C$

Slope of $B P$

Equation of side $A C$

Solving for Vertex $C (α, β)$

Calculate $α + β$

Circumcenter of $△ P A B$

First Equation for $(h, k)$

Second Equation for $(h, k)$

Solving for $h$ and $k$

Final Calculation

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

Analyzing the Setup

The Lines of Altitude

The Intersection of Destiny

The Circumcenter of $△ P A B$

Final Calculation

Similar Questions

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Select Answer:

Visualized Solution

Visualize the Given Points

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

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

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

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

Solving for Vertex .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C(α,β)

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

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

First Equation for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (h,k)

Second Equation for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (h,k)

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

Final Calculation

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

Analyzing the Setup

The Lines of Altitude

The Intersection of Destiny

The Circumcenter of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } △PAB

Final Calculation

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Slope of $A P$

Equation of side $B C$

Slope of $B P$

Equation of side $A C$

Solving for Vertex $C (α, β)$

Calculate $α + β$

Circumcenter of $△ P A B$

First Equation for $(h, k)$

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