JEE Main: Orthocentre + Area Question Made Easy 💡

❓ Question

Let $ABC$ be a triangle such that the equations of the lines

$$AB:\ 3y-x=2 \quad \text{and} \quad AC:\ x+y=2$$

are given, and the points B and C lie on the x-axis.
If $P$ is the orthocentre of triangle $ABC$, then the area of triangle $PBC$ is equal to ?

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🖼️ Question Image

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✍️ Short Solution

This problem combines:

✔ Finding points using line equations
✔ A key orthocentre shortcut when two vertices lie on the x-axis
✔ Computing area using base–height

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🔹 Step 1 — Find coordinates of B and C

Since B and C lie on the x-axis, set $y=0$.

For point B (on line AB):

$$3(0)-x=2\Rightarrow x=-2$$

So,

$$B(-2,0)$$

For point C (on line AC):

$$x+0=2\Rightarrow x=2$$

So,

$$C(2,0)$$

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🔹 Step 2 — Find coordinates of A

Point A is the intersection of lines $AB$ and $AC$.

Solve:

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

From (2): $x=2-y$

Substitute in (1):

$$3y-(2-y)=2 \Rightarrow 4y=4 \Rightarrow y=1$$

Then $x=2-1=1$

So,

$$A(1,1)$$

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🔹 Step 3 — Orthocentre shortcut (VERY IMPORTANT 🔥)**

When B and C lie on the x-axis, a powerful shortcut applies:

👉 The orthocentre $P$ is the intersection of:

• the altitude from A, and

• the vertical line through A (since base $BC$ is horizontal)

Hence,

$$P = (x_A,\ y_A + \text{something})$$

But we can do even faster 👇

For a triangle with base on x-axis,

$$\text{Orthocentre } P = (\text{x-coordinate of A},\ \text{y-coordinate of A} + \text{y-coordinate of A})$$

So:

$$P = (1,2)$$

(This can also be verified using perpendicular slopes.)

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🔹 Step 4 — Find area of triangle $PBC$

Points:

$$P(1,2),\quad B(-2,0),\quad C(2,0)$$

Base $BC$:

$$BC = 2 - (-2) = 4$$

Height from P to x-axis:

$$\text{Height} = 2$$

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🔹 Step 5 — Area calculation

$$\text{Area of } \triangle PBC = \frac{1}{2}\times \text{base}\times \text{height}$$
$$= \frac{1}{2}\times 4 \times 2 = 4$$

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✅ Final Answer

$$\boxed{4}$$