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If 2 and 3 are intercepts of a line $L=0$, then the distance of $L \equiv 0$ from the origin is
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Verified Answer
The correct answer is:
$\frac{6}{\sqrt{13}}$
If 2,3 are intercepts of a dine $L=0$, then
$$
\begin{aligned}
& x \text {-intercept }=2=a \\
& y \text {-intercept }=3=b \\
& \text { Equation of line is } \frac{x}{a}+\frac{y}{b}=1 \\
& \Rightarrow \quad \frac{x}{2}+\frac{y}{3}=1 \\
& \Rightarrow \quad 3 x+2 y-6=0
\end{aligned}
$$
$\therefore$ Required distance from origin is given by
$$
=\left|\frac{3(0)+2(0)-6}{\sqrt{3^2+2^2}}\right|=\frac{|-6|}{\sqrt{13}}=\frac{6}{\sqrt{13}}
$$
$\therefore$ Distance $=\frac{6}{\sqrt{13}}$ units
$$
\begin{aligned}
& x \text {-intercept }=2=a \\
& y \text {-intercept }=3=b \\
& \text { Equation of line is } \frac{x}{a}+\frac{y}{b}=1 \\
& \Rightarrow \quad \frac{x}{2}+\frac{y}{3}=1 \\
& \Rightarrow \quad 3 x+2 y-6=0
\end{aligned}
$$
$\therefore$ Required distance from origin is given by
$$
=\left|\frac{3(0)+2(0)-6}{\sqrt{3^2+2^2}}\right|=\frac{|-6|}{\sqrt{13}}=\frac{6}{\sqrt{13}}
$$
$\therefore$ Distance $=\frac{6}{\sqrt{13}}$ units
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