The point on the straight line $3x+4y = 8,$ which is closest to the origin is :

$\left(\frac{24}{25}, \frac{32}{25}\right)$

The correct answer is Option (2) → $\left(\frac{24}{25}, \frac{32}{25}\right)$

Perpendicular distance from a point $(x_0,y_0)$ is,

Distance = $\frac{|(Ax_0+By_0+C)|}{\sqrt{A^2+B^2}}$  line: $Ax+By+C$

$x=x_0-A\frac{Ax_0+By_0+C}{A^2+B^2},y=y_0-B\frac{Ax_0+By_0+C}{A^2+B^2}$

$Ax_0+By_0+C=3.0+4.0-8=-8$

$A^2+B^2=3^2+4^2=25$

$∴x=0.3(\frac{-8}{25})=\frac{24}{25}$

$y=0-4(\frac{-8}{25})=\frac{32}{25}$