JEE Mathematics 2025 Question with Solution

Given: The lines $AB$ and $AC$ are $3y - x = 2$ and $x + y = 2$ respectively, and $B$ and $C$ lie on the x-axis.

Find: The area of triangle $PBC$ where $P$ is the orthocentre of triangle $ABC$.

First, find the coordinates of $B$ and $C$ by putting $y = 0$.

For $AB:$

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

So, $B(-2,0)$.

For $AC:$

$$x + 0 = 2$$ $$x = 2$$

So, $C(2,0)$.

Now find point $A$ as the intersection of the two lines:

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

From $x + y = 2$, we get

$$x = 2 - y$$

Substitute into the first equation:

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

Hence,

$$x = 1$$

So, $A(1,1)$.

Since $BC$ lies on the x-axis, the altitude from $A$ is perpendicular to $BC$, so its equation is

$$x = 1$$

The slope of $AC$ is $-1$, so the altitude from $B$ has slope $1$. Passing through $B(-2,0)$, its equation is

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

Intersecting this with $x = 1$ gives

$$y = 3$$

Hence, the orthocentre is $P(1,3)$.

Now the base $BC = 2 - (-2) = 4$ and the perpendicular distance of $P$ from the x-axis is $3$. Therefore,

$$\text{Area of } \triangle PBC = \frac{1}{2} \times 4 \times 3 = 6$$

Therefore, the correct option is D.

Using the coordinate area formula with $P(1,3)$, $B(-2,0)$ and $C(2,0)$:

$$\text{Area} = \frac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|$$

Substitute the coordinates:

$$\text{Area} = \frac{1}{2}\left|1(0-0)+(-2)(0-3)+2(3-0)\right|$$ $$\text{Area} = \frac{1}{2}\left|0+6+6\right|$$ $$\text{Area} = \frac{1}{2} \times 12 = 6$$

So, the area of triangle $PBC$ is $6$.

The first extracted solution contains an internal discrepancy where it computes $8$ during substitution but still concludes $6$. The second approach is consistent and confirms the correct option as D.