Question

(a,b) is the point of concurrency of the lines \(x-3y+3=0\), \(Kx+y+k=0\) and \(2x+y-8=0\). If the perpendicular distance from the origin to the line \(L: ax-by+2k=0\) is \(p\), then the perpendicular distance from the point \((2,3)\) to \(L=0\) is:

• A. \(\frac{p}{2}\)
• B. \(p\)
• C. \(2p\)
• D. \(3p\)

Hint:

Check calculations for consistency and ensure that all geometric properties are considered, including the concurrency point calculation.

Answer 1

The Correct Option is B

We are given three lines:
1. \(x - 3y + 3 = 0\),
2. \(Kx + y + k = 0\),
3. \(2x + y - 8 = 0\).
The point \((a, b)\) is the point of concurrency of these three lines. We are also given a line \(L: ax - by + 2k = 0\), and the perpendicular distance from the origin to \(L\) is \(p\). We are asked to find the perpendicular distance from the point \((2, 3)\) to \(L\). Step 1: Find the point of concurrency \((a, b)\) For the three lines to be concurrent, they must all intersect at the same point \((a, b)\). This means \((a, b)\) satisfies all three equations: 1. \(a - 3b + 3 = 0\),
2. \(Ka + b + k = 0\),
3. \(2a + b - 8 = 0\).
From the third equation, solve for \(b\): \[ 2a + b - 8 = 0 \implies b = 8 - 2a \] Substitute \(b = 8 - 2a\) into the first equation: \[ a - 3(8 - 2a) + 3 = 0 \] \[ a - 24 + 6a + 3 = 0 \] \[ 7a - 21 = 0 \implies a = 3 \] Now substitute \(a = 3\) into \(b = 8 - 2a\): \[ b = 8 - 2(3) = 2 \] So, the point of concurrency is \((a, b) = (3, 2)\). Step 2: Find \(k\) using the second equation Substitute \((a, b) = (3, 2)\) into the second equation: \[ K(3) + 2 + k = 0 \] \[ 3K + 2 + k = 0 \] \[ 4K + 2 = 0 \implies K = -\frac{1}{2} \] Thus, \(k = K = -\frac{1}{2}\). Step 3: Write the equation of line \(L\) The line \(L\) is given by: \[ ax - by + 2k = 0 \] Substitute \(a = 3\), \(b = 2\), and \(k = -\frac{1}{2}\): \[ 3x - 2y + 2\left(-\frac{1}{2}\right) = 0 \] \[ 3x - 2y - 1 = 0 \] So, the equation of \(L\) is \(3x - 2y - 1 = 0\). Step 4: Find the perpendicular distance from the origin to \(L\) The perpendicular distance from a point \((x_0, y_0)\) to a line \(Ax + By + C = 0\) is given by: \[ \text{Distance} = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \] For the origin \((0, 0)\) and the line \(L: 3x - 2y - 1 = 0\), the distance \(p\) is: \[ p = \frac{|3(0) - 2(0) - 1|}{\sqrt{3^2 + (-2)^2}} = \frac{1}{\sqrt{9 + 4}} = \frac{1}{\sqrt{13}} \] Step 5: Find the perpendicular distance from \((2, 3)\) to \(L\) Using the same formula, the distance from \((2, 3)\) to \(L: 3x - 2y - 1 = 0\) is: \[ \text{Distance} = \frac{|3(2) - 2(3) - 1|}{\sqrt{3^2 + (-2)^2}} = \frac{|6 - 6 - 1|}{\sqrt{13}} = \frac{1}{\sqrt{13}} \] Thus, the distance from \((2, 3)\) to \(L\) is \(p\). Final Answer: \(\boxed{2}\)