Question

If $p$is the length of the perpendicular from the origin to the line, whose intercepts with the coordinate axes are $\frac{1}{3}$ and $\frac{1}{4}$ then the value of $p$ is

• A. $\frac{3}{5}$
• B. $\frac{1}{12}$
• C. $\frac{1}{5}$
• D. $5$

Answer 1

The correct option is C

$\frac{1}{5}$

Explanation for the correct options:

Step 1: Apply coordinate identities

The intercepts with the coordinate axes are $\frac{1}{3}$ and $\frac{1}{4}$.

So, the line passes through $\left(\frac{1}{3},0\right)$ and $\left(0,\frac{1}{4}\right)$

The slope of the line = $\left(\frac{{\displaystyle \frac{1}{4}-0}}{0-{\displaystyle \frac{1}{3}}}\right)=-\frac{3}{4}$ and the y-intercept is $\frac{1}{4}$
Step 2: Put this value in the below equation

The equation of the line is : $(y-y_1)=m(x-x_1)$

$y-0=-\frac{3}{4}\left(x-\frac{1}{3}\right)$

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

Simplifying this equation we get,

$\therefore 3x+4y=1$

Step 3: The length of the perpendicular from the origin,

$p=\pm \left(\frac{a{x}_1+b{y}_1+c}{\sqrt{a^2+b^2}}\right)$, where $p$ is the length and $a=3,b=4,x_1=0,y_1=0,\text{and}c=1$

$\begin{array}{rcl}\therefore \mathrm{P}& =& \frac{3\times 0-4\times 0+1}{\sqrt{3^2+4^2}}\\ & =& \frac{1}{\sqrt{9+16}}\\ & =& \frac{1}{\sqrt{25}}\\ & =& \frac{1}{5}\end{array}$

Hence option (3) is correct.