Let ABC be the triangle such that the equations of lines AB and AC be 3y - x = 2 and x + y = 2, resp

Q: Wait, why do we reject the negative root or why can't we just use the general formula for the orthocenter?

Great question! While there are general formulas for the orthocenter, they are often cumbersome and prone to calculation errors. In JEE Advanced, the beauty lies in exploiting the specific geometry of the problem. Here, because the base lies on the $x$-axis, the geometry is 'clean.' Always look for the simplest path—finding the intersection of two lines is much safer and faster than memorizing complex coordinate formulas.

Q: Why is the altitude from $A$ a vertical line? I thought altitudes were always slanted.

That is a common misconception! An altitude is simply a line segment from a vertex perpendicular to the opposite side. If the opposite side (the base $BC$) is horizontal (lying on the $x$-axis), the only line perpendicular to it is a vertical line. It is not about the triangle being slanted; it is about the orientation of the base relative to the coordinate axes.

Q: Is there a shortcut to finding the area of $\triangle PBC$ without finding the orthocenter $P$?

Not really. The area of $\triangle PBC$ is fundamentally tied to the height of $P$ above the base $BC$. Since $P$ is the orthocenter, its position is unique to the triangle $ABC$. You must find $P$ to determine its $y$-coordinate, which serves as the height of $\triangle PBC$. Embrace the process—calculating the intersection of two lines is a fundamental skill that will serve you well across all of coordinate geometry.

Question

Let $ABC$ be the triangle such that the equations of lines $AB$ and $AC$ be $3y - x = 2$ and $x + y = 2$, respectively, and the points $B$ and $C$ lie on $x$-axis. If $P$ is the orthocentre of the triangle $ABC$, then the area of the triangle $PBC$ is equal to

Options:

(A) 4

(B) 10

(C) 8

(D) 6

Sigma Percentile · Accepted Answer

### The Geometry of the Orthocenter: A Journey Through Coordinate Space

Welcome, my dear student. Today, we are not just solving a problem; we are embarking on a journey through the elegant landscape of coordinate geometry.

Often, when you see a problem involving an orthocenter, your mind might race to complex formulas involving slopes and determinants. But I want you to pause. Take a breath. Let's look at the geometry first.

The beauty of JEE Advanced problems lies in the fact that they often reward the student who stops to visualize before they start to calculate.

### Phase 1: Defining the Triangle

We are given two lines, $AB: 3y - x = 2$ and $AC: x + y = 2$. We are also told that points $B$ and $C$ lie on the $x$-axis.

If a point lies on the $x$-axis, its $y$-coordinate is zero. It is as simple as that!

Let's find the base of our triangle. For line $AB$, setting $y = 0$ gives us $-x = 2$, so $x = -2$. Thus, point $B$ is at $(-2, 0)$.

For line $AC$, setting $y = 0$ gives $x = 2$. Thus, point $C$ is at $(2, 0)$.

We have our base $BC$ lying perfectly on the $x$-axis, stretching from $-2$ to $2$. The length of this base is $2 - (-2) = 4$ units.

Now, what about vertex $A$? It is the intersection of our two lines. We solve the system:

$$3y - x = 2$$
$$x + y = 2$$

From the second equation, $x = 2 - y$. Substituting this into the first, we get $3y - (2 - y) = 2$, which simplifies to $4y = 4$, so $y = 1$.

Consequently, $x = 1$. Vertex $A$ is at $(1, 1)$. Our triangle is now fully defined in the coordinate plane.

### Phase 2: The Hunt for the Orthocenter

Now, we seek the orthocenter $P$. Remember, the orthocenter is the intersection of the altitudes. We need the equations of two altitudes.

Let's start with the altitude from $A(1, 1)$ to the base $BC$. Since $BC$ lies on the $x$-axis (a horizontal line), the altitude from $A$ must be a vertical line.

A vertical line has a constant $x$-coordinate. Since it passes through $A(1, 1)$, the equation is simply $x = 1$.

Next, we need the altitude from $B(-2, 0)$ to the side $AC$. The line $AC$ is $x + y = 2$, which we can rewrite as $y = -x + 2$.

The slope of $AC$ is $-1$. The altitude must be perpendicular to $AC$, so its slope must be the negative reciprocal of $-1$, which is $+1$.

Using the point-slope form for a line passing through $B(-2, 0)$ with slope $m = 1$:

$$y - 0 = 1(x - (-2))$$
$$y = x + 2$$

### Phase 3: The Final Area

We have our two altitudes: $x = 1$ and $y = x + 2$. Their intersection is the orthocenter $P$.

Substituting $x = 1$ into the second equation, we get $y = 1 + 2 = 3$. So, the orthocenter $P$ is at $(1, 3)$.

Finally, we need the area of $	riangle PBC$. The vertices are $P(1, 3)$, $B(-2, 0)$, and $C(2, 0)$.

The base $BC$ is on the $x$-axis with length $4$. The height of this triangle is the perpendicular distance from $P$ to the $x$-axis, which is simply the $y$-coordinate of $P$, which is $3$.

The area is given by the classic formula:

$$	ext{Area} = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$$
$$	ext{Area} = \frac{1}{2} 	imes 4 	imes 3 = 6$$

There we have it! **Six square units**. Notice how we didn't need to memorize any complex orthocenter formulas. We simply used the geometric properties of the triangle. Keep this mindset, and you will conquer any coordinate geometry problem that comes your way.

Select Answer:

Visualized Solution

Visualizing the Problem

Finding Points $B$ and $C$

Coordinates of $B$ and $C$

Finding Vertex $A$

Coordinates of $A$

The Orthocenter $P$

Altitude from $A$

Equation of Altitude from $A$

Altitude from $B$

Equation of Altitude from $B$

Finding Orthocenter $P$

Triangle $P B C$

Base and Height of $△ P B C$

Final Area Calculation

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

The Geometry of the Orthocenter

Phase 1

Phase 2

Phase 3

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Select Answer:

Visualized Solution

Visualizing the Problem

Finding Points .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

Coordinates of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

Finding Vertex .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

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

The Orthocenter .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Altitude from .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

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

Altitude from .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B

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

Finding Orthocenter .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

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

Base and Height of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } △PBC