Question

Find the perpendicular distance from the origin to the line x+3y+8=0
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Answer 1

Given ,

The line is x + 3y + 8 = 0

We know that , the distance of a point from a line is given by

$\sf \fbox{D = \frac{ |Ax_{1} + By_{1} + C| }{ \sqrt{ {(A)}^{2} + {(B)}^{2} } } }$

Thus ,

$\sf \mapsto D = \frac{ |1(0) + 3(0) + 8| }{ \sqrt{ {(1)}^{2} + {(3)}^{2} } } \\ \\ \sf \mapsto D = \frac{8}{ \sqrt{10} } \: \: units$

$\sf \therefore{ \underline{The \: perpendicular \: distance \: is \: \frac{8}{\sqrt{10}} \: units}}$