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3 and 5 are intercepts of a line $L=0$, then the distance of $L=0$ from $(3,7)$ is
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Verified Answer
The correct answer is:
$\frac{21}{\sqrt{34}}$
If 3 and 5 are intercepts of a line $L=0$, then
$x$-intercept $=a=3$
$y$-intercept $=b=5$
Equation of line is
$\frac{x}{3}+\frac{y}{5}=1 \Rightarrow 5 x+3 y-15=0$
$\therefore$ Required distance
$=\left|\frac{5(3)+3(7)-15}{\sqrt{5^2+3^2}}\right|=\frac{21}{\sqrt{34}}$
$x$-intercept $=a=3$
$y$-intercept $=b=5$
Equation of line is
$\frac{x}{3}+\frac{y}{5}=1 \Rightarrow 5 x+3 y-15=0$
$\therefore$ Required distance
$=\left|\frac{5(3)+3(7)-15}{\sqrt{5^2+3^2}}\right|=\frac{21}{\sqrt{34}}$
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