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Maths Question 11 – JEE-MAIN 2025

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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

Start by finding the coordinates of the vertices of triangle ABC, especially noting that B and C lie on the x-axis.

Step 1: Determine the coordinates of vertices A, B, and C.✦ Active

The vertex A is the intersection of lines 3y−x=2 and x+y=2. Solving these equations yields A=(1,1).

The vertex B is the intersection of line AB (3y−x=2) with the x-axis (y=0). Substituting y=0 gives x=−2, so B=(−2,0).

The vertex C is the intersection of line AC (x+y=2) with the x-axis (y=0). Substituting y=0 gives x=2, so C=(2,0).

Step 2: Find the coordinates of the orthocentre P.○ Expand

Since BC lies on the x-axis, the altitude from A to BC is a vertical line x=xA. Thus, the equation of the altitude from A is x=1.

The slope of line AB is mAB=13. The altitude from C to AB will have a slope mCE=−3. Using point C(2,0), the equation of this altitude is y−0=−3(x−2), which simplifies to y=−3x+6.

The orthocentre P is the intersection of x=1 and y=−3x+6. Substituting x=1 into the second equation gives y=−3(1)+6=3. Therefore, P=(1,3).

Step 3: Calculate the area of triangle PBC.○ Expand

The vertices of triangle PBC are P=(1,3), B=(−2,0), and C=(2,0).

The base BC lies on the x-axis. Its length is BC=|2−(−2)|=4.

The height of the triangle with respect to base BC is the perpendicular distance from P to the x-axis, which is the absolute value of the y-coordinate of P. Height h=|3|=3.

The area of triangle PBC is 12×base×height=12×4×3=6.

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