Proving regarding Position vectors

Question

How do I prove that the position vector of the point on a straight line closest to the origin is perpendicular to the line?

Answer 1

HINT

By a symmetry argument, indicating by $Q$ the intersection point of the line with the perpendicular from the origin, for $2$ symmetric points on the line $P'_1,P''_1$ with respect to $Q$ we have that $d_1=OP'_1=OP''_1$. Since for $P'_2,P''_2$ closer to $Q$ we have $d_2=OP'_2=OP''_2<d_1$, the minimum is reached when $P\equiv Q$.

By calculus

• line equation is: $ax+by+c=0$

• point $P_0(x_0,y_0)$ on the line such that $ax_0+by_0+c=0$

• distance form the origin to minimize $d=OP=\sqrt{x_0^2+y_0^2}$