Question

The points on the $x\text{-axis whose perpendicular distance from the line } \frac{x}{3} + \frac{y}{4} = 1 \text{ is 4 units are}$

• A. \( (8,0) \) and \( (-2,0) \)
• B. \( (-8,0) \) and \( (-2,0) \)
• C. \( (8,0) \) and \( (2,0) \)
• D. \( (-8,0) \) and \( (2,0) \)

Hint:

When calculating the perpendicular distance from a point to a line, remember to use the formula \( d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \) and rewrite the line in standard form.

Answer 1

The Correct Option is A

The equation of the given line is: \[ \frac{x}{3} + \frac{y}{4} = 1 \] We are required to find the points on the x-axis whose perpendicular distance from the given line is 4 units. The general formula to calculate the perpendicular distance of a point \( (x_1, y_1) \) from a line \( Ax + By + C = 0 \) is: \[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \] For the given equation \( \frac{x}{3} + \frac{y}{4} = 1 \), we can multiply both sides by 12 to rewrite it in standard form: \[ 4x + 3y - 12 = 0 \] Now, we can apply the distance formula. Let \( (x_1, y_1) = (x, 0) \) since we are considering points on the x-axis, and we need the perpendicular distance to be 4: \[ 4x + 3(0) - 12 = 4 \times \sqrt{4^2 + 3^2} \] Solving this equation: \[ \frac{|4x - 12|}{5} = 4 \] Multiply both sides by 5: \[ |4x - 12| = 20 \] This gives two cases: 1) \( 4x - 12 = 20 \) \[ 4x = 32 \quad \Rightarrow \quad x = 8 \] 2) \( 4x - 12 = -20 \) \[ 4x = -8 \quad \Rightarrow \quad x = -2 \] Thus, the points on the x-axis are \( (8,0) \) and \( (-2,0) \). Hence, the correct answer is (A).