JEE Main 2025 Apr 7 Shift 1, Mathematics Q20: Points and Straight Lines

JEE Main2025Apr 7, Shift 1Mathematics

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

A
$4$
B
$10$
C
$8$
D
$6$

Solution

Vertices. Vertex $A$ is the intersection of $3 y - x = 2$ and $x + y = 2$ . Solving: from the second $x = 2 - y$ ; substitute: $3 y - (2 - y) = 2 \Rightarrow 4 y = 4 \Rightarrow y = 1$ , $x = 1$ . So $A = (1, 1)$ .

$B$ is on the x-axis and on line $A B$ : put $y = 0$ in $3 y - x = 2$ : $x = - 2$ . So $B = (- 2, 0)$ .

$C$ is on the x-axis and on line $A C$ : put $y = 0$ in $x + y = 2$ : $x = 2$ . So $C = (2, 0)$ .

Orthocentre $P$ . Since $B C$ lies on the x-axis, the altitude from $A$ is the vertical line $x = 1$ .

The altitude from $B$ is perpendicular to $A C$ . Slope of $A C$ : $x + y = 2$ has slope $- 1$ , so the altitude has slope $1$ and passes through $B = (- 2, 0)$ : $y = x + 2$ .

Intersection: $x = 1 \Rightarrow y = 3$ . So $P = (1, 3)$ .

Area of $△ P B C$ . Base $B C$ lies on the x-axis with length $∣2 - (- 2) ∣ = 4$ . Height from $P$ to the x-axis is $∣3∣ = 3$ .

$Area = \frac{1}{2} \cdot 4 \cdot 3 = 6$

$∴$ Option (4).

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JEE Main 2025 Apr 7 Shift 1, Mathematics Q20: Points and Straight Lines

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