Let's solve the problem step-by-step: **Problem Restatement**: *Use mathematical optimization to prove that the shortest distance from a fixed point \(A\) to any point on a given line is the length of the line segment joining \(A\) to the line and is perpendicular to the line.* --- ### **Step 1: Setup** Let \(A = (x_0, y_0)\) be the fixed point. Let the given line be: \[ ax + by + c = 0 \] Let \(P = (x, y)\) be any point on the line. --- ### **Step 2: Write Distance Formula** The distance from \(A\) to \(P\) is: \[ d = \sqrt{(x - x_0)^2 + (y - y_0)^2} \] Our goal is to **minimize** \(d\), subject to the constraint that \(P\) lies on the line: \[ ax + by + c = 0 \] --- ### **Step 3: Use Lagrange Multipliers** Define the Lagrangian: \[ \mathcal{L}(x, y, \lambda) = (x - x_0)^2 + (y - y_0)^2 + \lambda (ax + by + c) \] We can minimize \(d^2\) instead of \(d\) (since the minimum occurs at the same point). --- ### **Step 4: Compute Partial Derivatives** Set partial derivatives to zero: 1. \(\frac{\partial \mathcal{L}}{\partial x} = 2(x - x_0) + \lambda a = 0\) 2. \(\frac{\partial \mathcal{L}}{\partial y} = 2(y - y_0) + \lambda b = 0\) 3. \(\frac{\partial \mathcal{L}}{\partial \lambda} = ax + by + c = 0\) --- ### **Step 5: Solve the System** From (1): \[ 2(x - x_0) + \lambda a = 0 \implies x = x_0 - \frac{\lambda a}{2} \] From (2): \[ 2(y - y_0) + \lambda b = 0 \implies y = y_0 - \frac{\lambda b}{2} \] Substitute \(x, y\) into the constraint (3): \[ a(x_0 - \frac{\lambda a}{2}) + b(y_0 - \frac{\lambda b}{2}) + c = 0 \] \[ a x_0 + b y_0 + c - \frac{\lambda}{2}(a^2 + b^2) = 0 \] \[ \frac{\lambda}{2}(a^2 + b^2) = a x_0 + b y_0 + c \] \[ \lambda = \frac{2(a x_0 + b y_0 + c)}{a^2 + b^2} \] --- ### **Step 6: Substitute \(\lambda\) Back** \[ x = x_0 - \frac{a}{2} \cdot \lambda = x_0 - \frac{a}{2} \cdot \frac{2(a x_0 + b y_0 + c)}{a^2 + b^2} = x_0 - \frac{a(a x_0 + b y_0 + c)}{a^2 + b^2} \] \[ y = y_0 - \frac{b}{2} \cdot \lambda = y_0 - \frac{b}{2} \cdot \frac{2(a x_0 + b y_0 + c)}{a^2 + b^2} = y_0 - \frac{b(a x_0 + b y_0 + c)}{a^2 + b^2} \] --- ### **Step 7: Compute the Minimum Distance** Use the point \((x, y)\) found above: \[ d_{\text{min}} = \sqrt{(x - x_0)^2 + (y - y_0)^2} \] Substitute: \[ x - x_0 = -\frac{a(a x_0 + b y_0 + c)}{a^2 + b^2} \] \[ y - y_0 = -\frac{b(a x_0 + b y_0 + c)}{a^2 + b^2} \] So, \[ d_{\text{min}} = \sqrt{ \left( \frac{a(a x_0 + b y_0 + c)}{a^2 + b^2} \right)^2 + \left( \frac{b(a x_0 + b y_0 + c)}{a^2 + b^2} \right)^2 } \] \[ = \frac{|a x_0 + b y_0 + c|}{\sqrt{a^2 + b^2}} \] --- ### **Conclusion** - This is the formula for the perpendicular distance from a point to a line. - The minimum (shortest) distance from \(A\) to the line is along the perpendicular. - The point on the line that achieves this distance is where the segment from \(A\) to the line is perpendicular. --- ### **Final Answer** **The shortest distance from a fixed point \(A\) to any point on a given line is the length of the line segment joining \(A\) to the line and is perpendicular to the line. This is given by** \[ d_{\text{min}} = \frac{|a x_0 + b y_0 + c|}{\sqrt{a^2 + b^2}} \] **using mathematical optimization.**