Chapter 6.2, Problem 67E

Solution

a.

To prove: The given points are the $y-\text{ and }x-\text{ intercepts}$ of $L$ .

Given information:

The line equation $L$ is $2x+5y=10$ .

  $A=(0,2)$

  $B=(5,0)$

Proof:

Let’s substitute the point $A=(0,2)$ in line equation.

  $\begin{array}{c}2x+5y=10\\ 2(0)+5(2)=10\\ 0+10=10\\ 10=10\end{array}$

Let’s substitute the point $B=(5,0)$ in line equation.

  $\begin{array}{c}2x+5y=10\\ 2(5)+5(0)=10\\ 10+0=10\\ 10=10\end{array}$

Therefore, the point $A\text{ and }B$ are the $y-\text{ and }x-\text{ intercepts}$ of $L$ .

b.

To find: The values of $w_1={\text{proj}}_{\overrightarrow{AB}}\overrightarrow{AP}\text{ and }{w}_2=\overrightarrow{AP}-{\text{proj}}_{\overrightarrow{AB}}\overrightarrow{AP}$ .

The value of,

  $w_1=\frac{5}{29}〈5,-2〉$

  $w_2=\frac{31}{29}〈2,5〉$

Given information:

The line equation $L$ is $2x+5y=10$ .

  $A=(0,2)$

  $B=(5,0)$

Formula used:

The projection of $u$ onto $v$ is $pro{j}_{v}u=\left(\frac{u\cdot v}{{\left|v\right|}^2}\right)v$

Calculation:

Using Head minus Method,

  $\begin{array}{c}\overrightarrow{AB}=〈5-0,0-2〉\\ =〈5,-2〉\\ \overrightarrow{AP}=〈3-0,7-2〉\\ =〈3,5〉\end{array}$

To find $w_1$:

  $\begin{array}{c}{w}_1=pro{j}_{\overrightarrow{AB}}\overrightarrow{AP}\\ =\left(\frac{\overrightarrow{AP}\cdot \overrightarrow{AB}}{|\overrightarrow{AB}{|}^2}\right)\overrightarrow{AB}\\ =\left(\frac{\langle 3,5\rangle \cdot \langle 5,-2\rangle }{|\langle 5,-2\rangle {|}^2}\right)\langle 5,-2\rangle \\ =\left(\frac{(3)(5)+(5)(-2)}{{\left(\sqrt{5^2+{(-2)}^2}\right)}^2}\right)\langle 5,-2\rangle \\ =\left(\frac{15-10}{{(\sqrt{29})}^2}\right)\langle 5,-2\rangle \\ w_1=\frac{5}{29}\langle 5,-2\rangle \end{array}$

To find $w_2$:

  $\begin{array}{c}{w}_2=\overrightarrow{AP}-pro{j}_{AB}\overrightarrow{AP}\\ =\langle 3,5\rangle -\frac{5}{29}\langle 5,-2\rangle \\ =\langle 3,5\rangle +〈-\frac{25}{29},\frac{10}{29}〉\\ =〈\frac{62}{29},\frac{155}{29}〉\\ w_2=\frac{31}{29}\langle 2,5\rangle \end{array}$

Therefore, the value of $w_1=\frac{5}{29}〈5,-2〉$ and $w_2=\frac{31}{29}〈2,5〉$ .

c.

To find: The distance from $P\text{ to L}$ .

The distance from $P\text{ to L}$ is, $\frac{31\sqrt{29}}{29}$ .

Given information:

The line equation $L$ is $2x+5y=10$ .

  $P=(3,7)$ .

Calculation:

A point and a line are separated by a distance that is perpendicular by definition.

Since w2 is parallel to the line L from the point P, it is the distance between P and L.

From part (b), it is known that

  $w_2=\frac{31}{29}〈2,5〉$

Thus,

  $\begin{array}{c}\left|w_2\right|=\frac{31}{29}\sqrt{2^2+5^2}\\ =\frac{31}{29}\sqrt{29}\\ \left|w_2\right|=\frac{31\sqrt{29}}{29}\end{array}$

Therefore, the distance from point to line is $\left|w_2\right|=\frac{31\sqrt{29}}{29}$ .

d.

To find: The formula for the distance from any point $P=(x_0,y_0)$ to $L$ .

The formula to find the distance from point $P\text{ to }L$ is, $\frac{\left|2x_0+5y_0-10\right|}{\sqrt{29}}$ .

Given information:

The line equation $L$ is $2x+5y=10$ .

  $A=(0,2)$

  $B=(5,0)$

Graph:

Considering the point $A\text{ and }B$ ,

  

Interpretation:

From the above graph, writing the vectors of $\overrightarrow{AB}\text{, }\overrightarrow{AP}$

  $\begin{array}{l}\overrightarrow{AB}=5i-2j\\ \overrightarrow{AP}=x_{0}i+(y_0-2)j\end{array}$

Now projecting the vector $\overrightarrow{AP}$ onto $\overrightarrow{AB}$ ,

  $\begin{array}{c}{\text{proj}}_{\overrightarrow{AB}}\overrightarrow{AP}=\overrightarrow{AR}\\ =\frac{\overrightarrow{AP}\times AB}{|\overrightarrow{AB}{|}^2}A\overrightarrow{B}\\ =\frac{\left[x^2+\left(y_0-2\right)\right]j\cdot (5j-2j)}{\left(\sqrt{{\left.5^2+{(-2)}^2\right)}^2}\right.}\left(5i-2j\right)\\ =\frac{5x_0-2\left(y_0-2\right)}{29}(5i-2j)\\ =\frac{5x_0-2y_0+4}{29}(5i-2j)\end{array}$

To find $\overrightarrow{RP}$

  $\begin{array}{c}\overrightarrow{RP}=\overrightarrow{AP}-\overrightarrow{AP}\\ =x_{0}i+\left(y_0-2\right)j-\frac{5x_0-2y_0+4}{29}(5i-2j)\\ =\frac{29x_0-25x_0+10y_0-20}{29}i+\frac{29y_0-58+10x_0-4y_0+8}{29}j\\ =\frac{4x_0+10y_0-20}{29}i+\frac{10x_0+25y_0-50}{29}j\end{array}$

Now let’s find the distance $RP$ ,

  $\begin{array}{c}RP=\sqrt{{\left(\frac{4x_0+10j_0-20}{29}\right)}^2+{\left(\frac{10x_0+25j_0-50}{29}\right)}^2}\\ =\sqrt{\frac{4x_0^2+25y_0^2+100+20x_0{y}_0-40x_0-100j_0}{29}}\\ =\frac{\left|2x_0+5y_0-10\right|}{\sqrt{29}}\end{array}$

Therefore, the distance from point $P\text{ to }L$ is, $\frac{\left|2x_0+5y_0-10\right|}{\sqrt{29}}$ .

e.

To find: The formula for the distance from any point $P=(x_0,y_0)$ to $L$ .

The formula to find the distance from point $P\text{ to }L$ is, $\frac{\left|a{x}_0+b{y}_0-c\right|}{\sqrt{a^2+b^2}}$ .

Given information:

The line equation $L$ is $ax+by=c$ .

  $A=(0,2)$

  $B=(5,0)$

Graph:

Considering the point $A\text{ and }B$ ,

  

Interpretation:

From the above graph, the points $A\text{ and }B$ are

  $\begin{array}{c}A=\left(0,\frac{c}{b}\right)\\ B=\left(\frac{c}{a},0\right)\end{array}$

Writing the vectors of $\overrightarrow{AB}\text{, }\overrightarrow{AP}$

  $\begin{array}{l}\overrightarrow{AB}=\frac{c}{a}i-\frac{c}{b}j\\ \overrightarrow{AP}=x_{0}i+(y_0-\frac{c}{b})j\end{array}$

Now projecting the vector $\overrightarrow{AP}$ onto $\overrightarrow{AB}$ ,

  $\begin{array}{c}pro{j}_{\overrightarrow{AB}}\overrightarrow{AP}=\overrightarrow{AR}\\ =\frac{\overrightarrow{AP}\times AB}{|\overrightarrow{AB}{|}^2}A\overrightarrow{B}\\ =\frac{\left[x_{0}i+\left(y_0-\frac{c}{b}\right)j\right]\left(\frac{c}{a}i-\frac{c}{b}j\right)}{{\left(\sqrt{{\left(\frac{c}{a}\right)}^2+{\left(-\frac{c}{b}\right)}^2}\right)}^2}\left(\frac{c}{a}i-\frac{c}{b}j\right)\\ =\frac{\frac{x_{0}c}{a}-\frac{y_{o}c}{b}+\frac{c^2}{b^2}}{\frac{c^2}{a^2}+\frac{c^2}{b^2}}\left(\frac{c}{a}i-\frac{c}{b}j\right)\end{array}$

To find $\overrightarrow{RP}$

  $\begin{array}{c}\overrightarrow{RP}=\overrightarrow{AP}-\overrightarrow{AP}\\ =\sqrt{{\left[\frac{a\left(a{x}_0+b{y}_0-c\right)}{a^2+b^2}\right]}^2+{\left[\frac{b\left(a{x}_0+b{y}_0-c\right)}{a^2+b^2}\right]}^2}\\ =\sqrt{\frac{{\left(a{x}_0+b{y}_0-c\right)}^2\left(a^2+b^2\right)}{{\left(a^2+b^2\right)}^2}}\\ =\frac{\left|a{x}_0+b{y}_0-c\right|}{\sqrt{a^2+b^2}}\end{array}$

Now let’s find the distance $RP$ ,

  $\begin{array}{c}RP=\sqrt{{\left(\frac{4x_0+10j_0-20}{29}\right)}^2+{\left(\frac{10x_0+25j_0-50}{29}\right)}^2}\\ =\sqrt{\frac{4x_0^2+25y_0^2+100+20x_0{y}_0-40x_0-100j_0}{29}}\\ =\frac{\left|2x_0+5y_0-10\right|}{\sqrt{29}}\end{array}$

Therefore, the distance from point $P\text{ to }L$ is, $\frac{\left|a{x}_0+b{y}_0-c\right|}{\sqrt{a^2+b^2}}$ .