Question

(a). Given the point $\left(2,1\right)$in the$\left(x,y\right)$ plane and the line $\mathbf{3}\mathit{x}\mathbf{+}\mathbf{2}\mathit{y}\mathbf{=}\mathbf{4}$, find the distance from the point to the line by using the method of Chapter 3, Section 5.

(b). Solve part (a) by writing a formula for the distance from $\left(2,1\right)$to$\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}$ and minimizing the distance (use Lagrange multipliers).

(c). Derive the formula

$\mathit{D}\mathbf{=}\left|\frac{a{x}_0+b{y}_0-c}{\sqrt{a^2+b^2}}\right|$

For the distance from $\left(x_0,y_0\right)$to $\mathit{a}\mathit{x}\mathbf{+}\mathit{b}\mathit{y}\mathbf{=}\mathit{c}$by the methods suggested in parts (a) and (b).

Short answer

(a). The distance from the point is$D=\frac{4}{\sqrt{13}}$ .

(b). The minimum distance is$1.18\text{units}$ .

(c). The derived formula is $PQ=\frac{\left|a{x}_1+b{y}_1+c\right|}{\sqrt{a^2+b^2}}$.

Step-by-step solution

(a) Find the distance from the point to the line.

The values derived from the given information are:

$$\begin{gathered} \left(x_0,y_0\right)=\left(2,1\right) \\ a=3 \\ b=2 \\ c=4 \end{gathered}$$

Now substitute the value in $D=\left|\frac{a{x}_0+b{y}_0-c}{\sqrt{a^2+b^2}}\right|$as:

$$\begin{gathered} D=\left|\frac{a{x}_0+b{y}_0-c}{\sqrt{a^2+b^2}}\right| \\ D=\left|\frac{3\times 2+2\times 1-4}{\sqrt{9+4}}\right| \\ D=\left|\frac{6+2-4}{\sqrt{13}}\right| \\ D=\frac{4}{\sqrt{13}} \end{gathered}$$

Therefore, the answer is$D=\frac{4}{\sqrt{13}}$ .

(b) Find the distance and minimizing the distance.

Finding the distance between the point $\left(2,1\right)$from$\left(x,y\right)$ as:

$\text{Distance}=\sqrt{{\left(2-x\right)}^2+{\left(1-y\right)}^2}$

Let $f\left(x,y\right)={\left(2-x\right)}^2+{\left(1-y\right)}^2$and$g\left(x,y\right)\Rightarrow 3x+2y-4=0$ .

We calculate minimum value of $f\left(x,y\right)$. We solve$\nabla f=\lambda \nabla g$ .

So:

$$\begin{gathered} f\left(x,y,\lambda \right)=f\left(x,y\right)+\lambda g\left(x,y\right) \\ f\left(x,y,\lambda \right)={\left(2-x\right)}^2+{\left(1-y\right)}^2+\lambda \left(3x+2y-4\right) \end{gathered}$$

Now finding$\frac{\partial f}{\partial x}$ as:

$$\begin{gathered} \frac{\partial f}{\partial x}=-2\left(2-x\right)+3\lambda \\ \frac{\partial f}{\partial x}=3\lambda +2x-4 \\ 0=3\lambda +2x-4 \\ \lambda =\frac{4-2x}{3} \end{gathered}$$

Now calculating the value of$\frac{\partial f}{\partial y}$ as:

$$\begin{gathered} \frac{\partial f}{\partial y}=-1\left(1-y\right)+2\lambda \\ \frac{\partial f}{\partial y}=2\lambda +y-1 \\ 0=2\lambda +y-1 \\ \lambda =\frac{1-y}{2} \end{gathered}$$

Now calculating the value of$\frac{\partial f}{\partial \lambda }$ as:

$$\begin{gathered} \frac{\partial f}{\partial y}=-1\left(1-y\right)+2\lambda \\ \frac{\partial f}{\partial \lambda }=3x+2y-4 \\ 0=3x+2y-4 \end{gathered}$$

Substituting the value of$\lambda$ equal as:

$$\begin{gathered} \frac{4-2x}{3}=\frac{1-y}{2} \\ 2\left(4-2x\right)=3\left(1-y\right) \\ 8-4x=3-3y \\ 4x-3y=5 \end{gathered}$$

Multiplying $3x+2y=4$equation by 4 and multiply $4x-3y=5$by 3as:

$$\begin{gathered} 12x+8y=16 \\ 12x-9y=15 \end{gathered}$$

Now subtract above equations as:

$$\begin{gathered} 12x+8y-\left(12x-9y\right)=16-15 \\ 12x+8y-12x+9y=1 \\ 17y=1 \\ y=\frac{1}{17} \end{gathered}$$

Now substitute the value of$y=\frac{1}{17}$ in$3x+2y=4$ as:

$$\begin{gathered} 3x+2y=4 \\ 3x+\frac{2}{17}=4 \\ 3x=\frac{66}{17} \\ x=\frac{22}{17} \end{gathered}$$

Substitute the value of x , y and in $f\left(x,y\right)={\left(2-x\right)}^2+{\left(1-y\right)}^2$as:

$$\begin{gathered} f\left(\frac{22}{17},\frac{1}{17}\right)={\left(2-\frac{22}{17}\right)}^2+{\left(1-\frac{1}{17}\right)}^2 \\ f\left(\frac{22}{17},\frac{1}{17}\right)={\left(\frac{34-22}{17}\right)}^2+{\left(\frac{17-1}{17}\right)}^2 \\ f\left(\frac{22}{17},\frac{1}{17}\right)={\left(\frac{12}{17}\right)}^2+{\left(\frac{16}{17}\right)}^2 \\ f\left(\frac{22}{17},\frac{1}{17}\right)=1.38 \end{gathered}$$

Now finding the distance is:

$$\begin{gathered} \text{Distance}=\sqrt{1.38} \\ \text{Distance}=1.18\text{units} \end{gathered}$$

Therefore, the answer is $1.18\text{units}$.

(c) Drive the formula.

The figure will be:

At point M,$x=0$ :

$$\begin{gathered} a\left(0\right)+by+c=0 \\ y=-\frac{c}{b} \end{gathered}$$

Now at point M, $y=0$:

$$\begin{gathered} ax+b\left(0\right)+c=0 \\ x=-\frac{c}{a} \end{gathered}$$

The area of the triangle MPN can be calculated as:

$$\begin{gathered} Are{a}_{\Delta MPN}=\frac{1}{2}\times \text{Base}\times \text{Height} \\ Are{a}_{\Delta MPN}=\frac{1}{2}\times \text{PQ}\times \text{MN} \\ PQ=\frac{2\times Are{a}_{\Delta MPN}}{\text{MN}}\left(1\right) \end{gathered}$$

The area of the triangle MPN can be calculated as:

$$\begin{gathered} Are{a}_{\Delta MPN}=\frac{1}{2}\times \left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right] \\ Are{a}_{\Delta MPN}=\frac{1}{2}\times \left[x_1\left(0+\frac{c}{b}\right)+\left(-\frac{c}{a}\right)\left(-\frac{c}{b}-y_1\right)+0\left(y_1-0\right)\right] \\ Are{a}_{\Delta MPN}=\frac{1}{2}\times \left[\frac{x_{1}c}{b}+\frac{y_{1}c}{a}+\frac{c^2}{ab}\right] \\ 2Are{a}_{\Delta MPN}=\frac{c}{ab}\left(a{x}_1+b{y}_1+c\right)\left(2\right) \end{gathered}$$

Now using the distance formula as:

$$\begin{gathered} MN=\sqrt{{\left(0+\frac{c}{a}\right)}^2+{\left(\frac{c}{b}-0\right)}^2} \\ MN=\frac{c}{ab}\sqrt{a^2+b^2}\left(3\right) \end{gathered}$$

Now from (1), (2), and (3) equations:

$PQ=\frac{\left|a{x}_1+b{y}_1+c\right|}{\sqrt{a^2+b^2}}$

Hence the formula derived is $PQ=\frac{\left|a{x}_1+b{y}_1+c\right|}{\sqrt{a^2+b^2}}$.