Question

find the distance of the origin from the line 7x+24y-50=0​

Answer 1

$\textbf{Concept used:}$

$\text{The perpendicular distance of the point}$

$\text{ (x_1,y_1) from the line ax+by+c=0 is}$

$\displaystyle\bf|\frac{ax_1+by_1+c}{\sqrt{a^2+b^2}}|$

$\text{Now,}$

$\text{The perpendicular distance of the point}$

$\text{(0,0) from the line 7x+24y-50=0 is}$

$\displaystyle\bf|\frac{ax_1+by_1+c}{\sqrt{a^2+b^2}}|$

$=\displaystyle|\frac{7x_1+24y_1-50}{\sqrt{7^2+24^2}}|$

$=\displaystyle|\frac{7(0)+24(0)-50}{\sqrt{49+576}}|$

$=\displaystyle|\frac{-50}{\sqrt{625}}|$

$=\displaystyle|\frac{-50}{25}|$

$=\displaystyle|-2|$

$=2\;\text{units}$

$\therefore\textbf{The distance of the (0,0) from the line 7x+24y-50=0 is 2 units}$