# Find the point on the line 5x+4y-4=0 which is closest to the point (-3,0).

## Question:

Find the point on the line {eq}\ 5x+4y-4=0 \ {/eq} which is closest to the point {eq}(-3,0) {/eq}.

## Minimum Distance:

The minimum distance between a line and a point can be found with the Lagrange Multipliers, the objective function must be that formed with the given point and the constraint is the line.

## Answer and Explanation:

We have,

The point: {eq}(-3,0) \\ {/eq}

{eq}d= \sqrt{ (x-x_0)^{2}+(y-y_0)^{2} } \\ d= \sqrt{ (x-3)^{2}+(y-0)^{2} } \\ d= \sqrt{ (x-3)^{2}+(y)^{2} } \\ d^{2}=(x-3)^{2}+(y)^{2} {/eq}

Our function is: {eq}f(x,y)=(x-3)^{2}+(y)^{2} \\ {/eq}

Using Lagrange multipliers:

We have the function

{eq}f(x,y)=(x-3)^{2}+(y)^{2} \\ {/eq}

Differentiating the function

{eq}f_x=2\,x-6 \\ f_y=2\,y \\ {/eq}

Constraint:

{eq}g_{x,y}=5x+4y-4 \\ {/eq}

Derivatives:

{eq}g_x=5 \\ g_y=4 \\ {/eq}

Now, get the following equation system:

{eq}f_{x}(x,y)= \lambda g_{x}(x,y) \\ \Rightarrow f_{x}(x,y)- \lambda g_{x}(x,y)=0 \\ {/eq}

{eq}\begin {align} \Rightarrow & -5\,\lambda+2\,x-6 =0 & (1) \\ \end {align} {/eq}

{eq}f_{y}(x,y)= \lambda g_{y}(x,y) \\ \Rightarrow f_{y}(x,y)- \lambda g_{y}(x,y) =0 \\ {/eq}

{eq}\begin {align} \Rightarrow & -4\,\lambda+2\,y =0 & (2) \\ \end {align} {/eq}

{eq}\begin {align} Constraint \, \, \Rightarrow & 5\,x+4\,y-4 =0 & (3) \\ \end {align} {/eq}

We perform the system of equations and find:

{eq}\begin{array} \; \; \text{(x,y)} \; & \lambda & (x,y, f(x,y)) \\ \hline ( \frac{68}{41}, -\frac{44}{41} ) & -\frac{22}{41} \; & (\frac{68}{41}, -\frac{44}{41}, \frac{121}{41} ) \\ \end{array} \\ {/eq}

The distance is:

{eq}d^{2}= \frac{121}{41} \\ d= \sqrt{ \frac{121}{41} } \\ d= 1.7179 \\ {/eq}

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from Math 104: Calculus

Chapter 9 / Lesson 3