Question

Let the point $P(\alpha, \beta)$ be at a unit distance from each of the two lines $L _1: 3 x -4 y +12=0$, and $L _2: 8 x +6 y +11=0$ If $P$ lies below $L _1$ and above $L _2$, then $100(\alpha+\beta)$ is equal to

• A. $-14$
• B. 42
• C. $-22$
• D. 14

Answer 1

The Correct Option is D

To determine the value of \(100(\alpha + \beta)\) for the point \(P(\alpha, \beta)\) that is at a unit distance from the given lines, we'll proceed with the following steps:

1. Identify the given lines and their equations:
   • Line \(L_1: 3x - 4y + 12 = 0\)
   • Line \(L_2: 8x + 6y + 11 = 0\)
2. Recall that the distance \(d\) from a point \((x_1, y_1)\) to a line \(Ax + By + C = 0\) is given by the formula: 
   \(d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}\)
3. Apply this formula to the point \(P(\alpha, \beta)\) for each line, setting the distance equal to 1. 
   For line \(L_1\): \(\frac{|3\alpha - 4\beta + 12|}{\sqrt{3^2 + (-4)^2}} = 1\) Simplifying, we get: \(|3\alpha - 4\beta + 12| = 5\)
4. For line \(L_2\): 
   \(\frac{|8\alpha + 6\beta + 11|}{\sqrt{8^2 + 6^2}} = 1\) Simplifying, we obtain: \(|8\alpha + 6\beta + 11| = 10\)
5. Now, split each absolute value equation into two linear equations: 
   For \(L_1\):
   • \(3\alpha - 4\beta + 12 = 5\) or \(3\alpha - 4\beta + 12 = -5\)
   • \(8\alpha + 6\beta + 11 = 10\) or \(8\alpha + 6\beta + 11 = -10\)
6. Solve these systems under the conditions stated: \(P\) lies below \(L_1\) and above \(L_2\).
   • For \(3\alpha - 4\beta + 12 = 5\), it simplifies to \(3\alpha - 4\beta = -7\).
   • For \(8\alpha + 6\beta + 11 = 10\), it simplifies to \(8\alpha + 6\beta = -1\).
7. Solve this simultaneous linear system:
   • Multiply the first equation by 2: \(6\alpha - 8\beta = -14\)
   • Add to the second equation: \(8\alpha + 6\beta = -1\)
   • Solve: \( \alpha = \frac{-1}{14} \) and \(\beta = \frac{1}{7} \)
8. Compute \(\alpha + \beta = \frac{-1}{14} + \frac{1}{7} = 0\) and, ultimately, \(100(\alpha + \beta) = 14\).

The correct answer is indeed 14.