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: Concept
The perpendicular distance $d$ from a point $(x_1, y_1)$ to a line $ax + by + c = 0$ is given by $d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$.

Step 2: Analysis
Given $d = 5$, point $(1, 2)$, and line $4x - 3y + k = 0$. $5 = \frac{|4(1) - 3(2) + k|}{\sqrt{4^2 + (-3)^2}} = \frac{|4 - 6 + k|}{5}$.

Step 3: Calculation
$25 = |k - 2| \implies k - 2 = 25$ or $k - 2 = -25$. $k = 27$ or $k = -23$. Final Answer: (A)