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. $\frac{5}{12}$ units
• D. $\frac{12}{5}$ units

Answer 1

The correct option is D

$\frac{12}{5}$ units

Explanation for the correct option:

Step 1. Find the distance of the point from a line:

Given,

x intercept $=3$

y intercept $=4$

Now, Intercept form of Equation is $\frac{\mathbf{x}}{\mathbf{a}}\mathbf{}\mathbf{+}\mathbf{}\frac{\mathbf{y}}{\mathbf{b}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}$

$\Rightarrow$ $\frac{x}{3}+\frac{y}{4}=1$

$\Rightarrow$ $4x+3y–12=0$

Step 2. Distance of a point $P(x_1,y_1)$ from a line $ax+by+c=0$ is $\mathit{d}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{|}\mathbf{a}{\mathbf{x}}_{\mathbf{1}}\mathbf{}\mathbf{+}\mathbf{}\mathbf{b}{\mathbf{y}}_{\mathbf{1}}\mathbf{}\mathbf{+}\mathbf{}\mathbf{c}\mathbf{|}}{\sqrt{\mathbf{(}{\mathbf{a}}^{\mathbf{2}}\mathbf{}\mathbf{+}\mathbf{}{\mathbf{b}}^{\mathbf{2}}\mathbf{)}}}$

Here the point is at the origin.

So, $P=(0,0)$

$a=4,b=3$ and $c=-12$

$\begin{array}{rcl}\therefore d& =& \frac{|0+0–12|}{\sqrt{(42+32)}}\\ & =& \frac{\mathbf{12}}{\mathbf{5}}\end{array}$

Hence, Option ‘D’ is Correct.