Animated Solution for Mathematics - Straight Lines: Let the point $P(\alpha ,\beta )$ be at a unit distance from each of the two lines $L_1:3x-4y+12=0$, and $L_2:8x+6y+11=0$. If $P$ lies below $L_1$ and above $L_2$, then $100(\alpha +\beta )$ is equal to

Imagine you are standing on a vast coordinate plane, a silent observer of two intersecting laser beams, $L_1$ and $L_2$. You are tasked with finding a specific point $P(\alpha ,\beta )$ that acts like a precise navigator, maintaining a strict, unwavering distance of exactly $1$ unit from both beams.

This is not just a problem of algebra; it is a problem of spatial awareness. We are looking for the intersection of two loci, where each locus is a pair of parallel lines.

To find this point, we must first master the tool of our trade: the perpendicular distance formula. For any line $Ax+By+C=0$, the distance $d$ from a point $(\alpha ,\beta )$ is given by:

The modulus bars here are the gatekeepers of our logic. They exist because distance is a scalar, always positive, while the expression $A\alpha +B\beta +C$ can be positive or negative depending on which side of the line the point resides.

This is where the magic of the 'Origin Test' comes in. By plugging $(0,0)$ into our line equations, we anchor our coordinate system. For $L_1:3x-4y+12=0$, the origin yields $12$, a positive value.

Since the origin is below $L_1$, any point below $L_1$ must satisfy the same sign condition. Thus, for our point $P$, the expression $3\alpha -4\beta +12$ must be positive, allowing us to drop the modulus bars with confidence.

With the modulus resolved, the geometry collapses into simple, elegant linear algebra. For $L_1$, we have:

We repeat this process for $L_2:8x+6y+11=0$. Testing the origin again, we find $11>0$, and since the origin is above $L_2$, the expression $8\alpha +6\beta +11$ must also be positive.

This gives us our second equation:

Now, we stand before a system of two linear equations: 1) $3\alpha -4\beta =-7$ 2) $8\alpha +6\beta =-1$

By multiplying the first by $3$ and the second by $2$, we orchestrate a beautiful cancellation of the $\beta$ terms. The result is $25\alpha =-23$, leading us to $\alpha =-\frac{23}{25}$. Substituting this back, we find $\beta =\frac{53}{50}$.

The journey concludes with a simple summation. We need $100(\alpha +\beta )$. Adding our values, we get:

Multiplying by $100$, we arrive at the final answer: 14.

You have navigated the coordinate plane, resolved the ambiguity of the modulus, and solved the system with precision. This is the essence of JEE Advanced mathematics—not just finding the number, but understanding the geometric soul of the problem.