Question

If 3 and 4 are intercepts of a line \( L = 0 \), then the distance of \( L = 0 \) from the origin is:

• A. 5 units
• B. 12 units
• C. \( \frac{5}{12} \) unit
• D. \( \frac{12}{5} \) units

Hint:

In intercept form, the equation of a line is \( \frac{x}{a} + \frac{y}{b} = 1 \), where \( a \) and \( b \) are the x and y intercepts, respectively.

Answer 1

The Correct Option is D

We are given that the intercepts of the line are 3 and 4. The equation of the line in the intercept form is: \[ \frac{x}{3} + \frac{y}{4} = 1 \] The distance \( D \) of a line \( Ax + By + C = 0 \) from the origin is given by the formula: \[ D = \frac{|C|}{\sqrt{A^2 + B^2}} \] In this case, we rewrite the line in the general form \( Ax + By + C = 0 \) as: \[ \frac{x}{3} + \frac{y}{4} - 1 = 0 \quad \Rightarrow \quad 4x + 3y - 12 = 0 \] Thus, \( A = 4, B = 3, C = -12 \). Using the formula for distance: \[ D = \frac{|(-12)|}{\sqrt{4^2 + 3^2}} = \frac{12}{\sqrt{16 + 9}} = \frac{12}{\sqrt{25}} = \frac{12}{5} \] Thus, the correct answer is \( \frac{12}{5} \) units.