Question

Find the nearest point on the line $3x+4y-1=0$ from the origin.

• A. $(\frac{3}{25},\frac{4}{25})$
• B. $(\frac{1}{4},\frac{1}{8})$
• C. $(\frac{1}{6},\frac{1}{9})$
• D. $(\frac{1}{7},\frac{1}{4})$

Answer 1

Answer: (A)

$(\frac{3}{25},\frac{4}{25})$

The nearest point will be the foot of the perpendicular from the origin

$3x+4y-1=0(x_0,y_0)=(0,0)\Rightarrow \frac{x-x_0}{3}=\frac{y-y_0}{4}=-\left(\frac{3x_0+4y_0-1}{{3^2+4}^2}\right)\Rightarrow \frac{x-0}{3}=\frac{y-0}{4}=\frac{1}{25}\Rightarrow x=\frac{3}{25},y=\frac{4}{25}$

Required ponit is $(\frac{3}{25},\frac{4}{25})$