Animated Solution for Mathematics - Straight Lines: If the line, $2x-y+3=0$ is at a distance $\frac{1}{\sqrt{5}}$ and $\frac{2}{\sqrt{5}}$ from the lines $4x-2y+\alpha =0$ and $6x-3y+\beta =0$, respectively, then the sum of all possible values of $\alpha$ and $\beta$ is

Welcome, future engineer. Today, we are not just solving an algebra problem; we are exploring the architecture of the Cartesian plane.

When you look at the equations $2x-y+3=0$, $4x-2y+\alpha =0$, and $6x-3y+\beta =0$, do not just see numbers and variables. See a family of parallel lines, marching across the coordinate plane, maintaining a constant, elegant distance from one another.

This problem is a classic test of your ability to normalize your perspective—a skill that is vital in both physics and engineering.

The Anchor and the Trap

Let us begin our journey by observing the anchor of our problem: the line $L_1$ defined by $2x-y+3=0$. Its slope is $2$.

Now, look at the second line, $L_2:4x-2y+\alpha =0$. If you look closely, you will see that the coefficients of $x$ and $y$ are exactly double those of $L_1$. This is the first realization: these lines are parallel.

However, here lies the trap. Many students rush to apply the distance formula immediately. But wait! The standard distance formula between two parallel lines, $Ax+By+C_1=0$ and $Ax+By+C_2=0$, is given by:

Notice that the coefficients $A$ and $B$ must be identical. In our case, $L_1$ has coefficients $(2,-1)$, while $L_2$ has $(4,-2)$.

We must normalize $L_2$. By dividing the entire equation $4x-2y+\alpha =0$ by $2$, we transform it into $2x-y+\frac{\alpha }{2}=0$. Now, and only now, are we speaking the same language.

The Modulus Mystery

With our equations normalized, we can apply the distance formula. We are given that the distance is $\frac{1}{\sqrt{5}}$. Substituting our values, we get:

Simplifying the denominator, $\sqrt{4+1}=\sqrt{5}$. The $\sqrt{5}$ terms cancel out beautifully, leaving us with the absolute value equation:

This is the moment where intuition meets algebra. Why the absolute value? Because geometry is symmetric.

A line at a distance of $\frac{1}{\sqrt{5}}$ from $L_1$ can exist on either side—above or below. The modulus captures both possibilities. Solving for $\frac{\alpha }{2}-3=1$ gives $\alpha =8$, and solving for $\frac{\alpha }{2}-3=-1$ gives $\alpha =4$. Both are valid, and both are necessary.

The Symmetry of $\beta$

We repeat this elegant process for the third line, $L_3:6x-3y+\beta =0$. Again, we normalize. Dividing by $3$, we obtain $2x-y+\frac{\beta }{3}=0$.

The problem states the distance from $L_1$ is $\frac{2}{\sqrt{5}}$. Applying our formula once more:

Canceling the $\sqrt{5}$ yields $|\frac{\beta }{3}-3|=2$. Just as before, we branch into two cases:

1. $\frac{\beta }{3}-3=2\Rightarrow \frac{\beta }{3}=5\Rightarrow \beta =15$ 2. $\frac{\beta }{3}-3=-2\Rightarrow \frac{\beta }{3}=1\Rightarrow \beta =3$

We have navigated the geometry, respected the normalization, and embraced the symmetry of the modulus. We have found two possible values for $\alpha$ ($4$ and $8$) and two possible values for $\beta$ ($3$ and $15$).

The final step is simply to sum these values:

Thirty. It is a clean, satisfying integer. Remember, in JEE Advanced, the math is rarely just about the final number; it is about the clarity of the path you took to get there. You have mastered the parallel line distance formula today. Carry this confidence into your next problem.