Question

If the distance of the line $4x-3y+k=0$ from the point (1, 2) is 5 units, then the values of k are

• A. 27, -23
• B. -27, 23
• C. 29, -24
• D. -29, 24
• E. -28, -25

Hint:

When removing the absolute value sign $|x| = a$, always consider both $+a$ and $-a$ cases.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
We need to find the value of a parameter 'k' in the equation of a line, given the perpendicular distance from a specific point to that line.
Step 2: Key Formula or Approach:
The formula for the perpendicular distance (d) from a point \( (x_1, y_1) \) to a line \( Ax + By + C = 0 \) is:
\[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \] Step 3: Detailed Explanation:
We are given:
The line: \( 4x - 3y + k = 0 \). So, \( A=4, B=-3, C=k \).
The point: \( (x_1, y_1) = (1, 2) \).
The distance: \( d = 5 \).
Substitute these values into the distance formula:
\[ 5 = \frac{|4(1) - 3(2) + k|}{\sqrt{4^2 + (-3)^2}} \] Now, we solve for k. First, simplify the expression:
\[ 5 = \frac{|4 - 6 + k|}{\sqrt{16 + 9}} \] \[ 5 = \frac{|k - 2|}{\sqrt{25}} \] \[ 5 = \frac{|k - 2|}{5} \] Multiply both sides by 5:
\[ 25 = |k - 2| \] This absolute value equation gives two possible linear equations:
Case 1: \( k - 2 = 25 \)
\[ k = 25 + 2 = 27 \] Case 2: \( k - 2 = -25 \)
\[ k = -25 + 2 = -23 \] The two possible values for k are 27 and -23.
Step 4: Final Answer:
The values of k are 27 and -23.