Let the Straight Lines AB: x + y − 2 = 0 and AC: 3y − x = 2 Intersect the x-axis at B and C Respectively. If P Is the Orthocentre of Triangle ABC, Then the Area of Triangle CPB Is

JEE Main 2025 — 7 April, Morning Shift. Previous Year Question.

Problem

Let the straight lines $AB: x+y-2=0$ and $AC: 3y-x=2$ intersect the x-axis at $B$ and $C$ respectively. If $P$ is the orthocentre of the triangle $ABC$, then the area of the triangle $CPB$ is:

Key insight. Since $B$ and $C$ both lie on the x-axis, triangle $CPB$ has its base $CB$ sitting directly on the x-axis — making its area trivial to compute once the orthocentre’s height above the x-axis is known: just half the base times that height.

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Approach

Find $B$ and $C$ by setting $y=0$ in each line’s equation. Find $A$ as the intersection of the two lines. Since $BC$ lies along the x-axis, the altitude from $A$ is simply the vertical line through $A$; find where this meets the altitude from $B$ (perpendicular to $AC$) to locate the orthocentre $P$. Then compute the area of $\triangle CPB$ using base $CB$ and the height of $P$ above the x-axis.

Solution

Step 1 — Find B (where AB meets the x-axis)

Setting $y=0$ in $x+y-2=0$: $x=2$, so $B=(2,0)$.

Step 2 — Find C (where AC meets the x-axis)

Setting $y=0$ in $3y-x=2$: $-x=2 \implies x=-2$, so $C=(-2,0)$.

Step 3 — Find A (intersection of AB and AC)

Solving $x+y=2$ and $3y-x=2$ simultaneously: adding gives $4y=4 \implies y=1$, and then $x=2-1=1$. So $A=(1,1)$.

Step 4 — Find the altitude from A

Since $BC$ lies along the x-axis, the altitude from $A$ (perpendicular to $BC$) is the vertical line:

$$x=1$$

Step 5 — Find the altitude from B

The altitude from $B$ is perpendicular to $AC$. Since $AC: 3y-x=2 \implies y=\dfrac{x+2}{3}$ has slope $\dfrac{1}{3}$, the altitude has slope $-3$, passing through $B(2,0)$:

$$y = -3(x-2) = -3x+6$$

Step 6 — Find the orthocentre P

Intersecting $x=1$ with $y=-3x+6$:

$$y = -3(1)+6 = 3 \implies P=(1,3)$$

Step 7 — Compute the area of triangle CPB

Since $C=(-2,0)$ and $B=(2,0)$ both lie on the x-axis, the base $CB$ has length:

$$CB = |2-(-2)| = 4$$

The height is the perpendicular distance from $P=(1,3)$ to the x-axis, which is simply its $y$-coordinate: $3$.

$$\text{Area} = \frac{1}{2}\times\text{base}\times\text{height} = \frac{1}{2}\times4\times3 = 6$$

Answer

Common mistakes

• Forgetting that $B$ and $C$ lying on the x-axis makes the height calculation trivial. Without noticing this, it’s tempting to use the general “half the magnitude of a cross product” formula for the triangle’s area — which also works, but is more computation than needed here.
• Mixing up which altitude is perpendicular to which side. The altitude from $B$ must be perpendicular to the opposite side $AC$ (not $AB$) — using the wrong side’s slope gives an incorrect orthocentre.

Practise next

• Lines $PQ: 2x+y-4=0$ and $PR: y-2x=6$ meet the x-axis at $Q$ and $R$ respectively. If $S$ is the orthocentre of $\triangle PQR$, find the area of $\triangle RSQ$, using the same approach.