The point on the straight line $3x+4y = 8,$ which is closest to the origin is :
Answer & explanation
Correct answer: option 2
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: $Ax+By+C$
$x=x_0-A\frac{Ax_0+By_0+C}{A^2+B^2},y=y_0-B\frac{Ax_0+By_0+C}{A^2+B^2}$
$Ax_0+By_0+C=3.0+4.0-8=-8$
$A^2+B^2=3^2+4^2=25$
$∴x=0.3(\frac{-8}{25})=\frac{24}{25}$
$y=0-4(\frac{-8}{25})=\frac{32}{25}$