Answer
Using Theorem 1, we show that the point $\left( {{x_0},{y_0}} \right)$ closest to the origin on the line $ax + by = c$ has coordinates
${x_0} = \frac{{ac}}{{{a^2} + {b^2}}}$, ${\ \ \ }$ ${y_0} = \frac{{bc}}{{{a^2} + {b^2}}}$
Work Step by Step
We are given the line $ax + by = c$. Our task is to minimize the distance $d = \sqrt {{x^2} + {y^2}} $ subject to the constraint $g\left( {x,y} \right) = ax + by - c = 0$.
Since finding the minimum distance $d$ is the same as finding the minimum square of the distance ${d^2}$, our task is to minimize $f\left( {x,y} \right) = {x^2} + {y^2}$ subject to the constraint $g\left( {x,y} \right) = ax + by - c = 0$.
Step 1. Write out the Lagrange equations
Using Theorem 1, the Lagrange condition $\nabla f = \lambda \nabla g$ yields
$\left( {2x,2y} \right) = \lambda \left( {a,b} \right)$
So, the Lagrange equations are
(1) ${\ \ \ \ }$ $2x = a\lambda $, ${\ \ \ }$ $2y = b\lambda $
Step 2. Solve for $\lambda$ in terms of $x$ and $y$
Since $\left( {0,0} \right)$ does not satisfy the constraint, we may assume that $x \ne 0$ and $y \ne 0$.
Thus, $\lambda = \frac{2}{a}x = \frac{2}{b}y$.
Step 3. Solve for $x$ and $y$ using the constraint
From Step 2 we obtain $y = \frac{b}{a}x$.
Substituting it in the constraint gives
$ax + b\left( {\frac{b}{a}x} \right) - c = 0$
$x\left( {a + \frac{{{b^2}}}{a}} \right) = c$
So, $x = \frac{{ac}}{{{a^2} + {b^2}}}$.
Using $y = \frac{b}{a}x$, we obtain the critical point: $\left( {{x_0},{y_0}} \right) = \left( {\frac{{ac}}{{{a^2} + {b^2}}},\frac{{bc}}{{{a^2} + {b^2}}}} \right)$. This critical point corresponds to the minimum of $f$ subject to the constraint $g\left( {x,y} \right) = ax + by - c = 0$.
Step 4. Calculate the critical values
We evaluate the minimum value $f\left( {\frac{{ac}}{{{a^2} + {b^2}}},\frac{{bc}}{{{a^2} + {b^2}}}} \right) = \frac{{{c^2}}}{{{a^2} + {b^2}}}$ (the minimum square of the distance).
Notice that there is no maximum value since there are points on the line that are arbitrarily far from the origin. Hence, the point $\left( {{x_0},{y_0}} \right)$ closest to the origin on the line $ax + by = c$ has coordinates
${x_0} = \frac{{ac}}{{{a^2} + {b^2}}}$, ${\ \ \ }$ ${y_0} = \frac{{bc}}{{{a^2} + {b^2}}}$
