Animated Solution for Mathematics - Straight Lines: Let two points be $A(1,-1)$ and $B(0,2)$. If a point $P(x^{\prime },y^{\prime })$ be such that the area of $\Delta PAB=5$ sq. units and it lies on the line, $3x+y-4\lambda =0$, then the value of $\lambda$ is :

Imagine you are standing on a coordinate plane with two fixed anchors, point $A(1,-1)$ and point $B(0,2)$. These two points define a rigid, unmoving base for a triangle.

A third point, $P(x^{\prime },y^{\prime })$, wanders across the plane. We constrain this wanderer such that the area of $\Delta PAB$ is always exactly $5$ square units.

To solve this, we use the coordinate area formula. For any triangle with vertices $(x_1,y_1)$, $(x_2,y_2)$, and $(x_3,y_3)$, the area is given by:

Plugging in our points $A(1,-1)$, $B(0,2)$, and $P(x^{\prime },y^{\prime })$, the expression simplifies because the $x$-coordinate of $B$ is $0$:

This collapses into the following equation:

We now have $|2-y^{\prime }-3x^{\prime }|=10$. The modulus sign is a gateway to two distinct realities.

When we strip away the absolute value, we must account for both the positive and negative possibilities:

These represent two parallel lines. Geometrically, if you keep the base $AB$ fixed and maintain a constant area, the height of the triangle must be constant. There are exactly two lines parallel to $AB$ that maintain this specific height.

We are given the constraint that $P$ lies on the line $3x+y-4\lambda =0$. This implies that the coordinates of $P$ must satisfy:

We have established that for the area to be $5$, $3x^{\prime }+y^{\prime }$ must be either $12$ or $-8$. Therefore, $4\lambda$ must equal $12$ or $-8$.

Solving these gives us:

The problem is a classic JEE favorite because it forces you to look past the raw algebra and see the parallel lines hidden in the equations. The final result for $\lambda$ is $3$ or $-2$.