Chapter 9.5, Problem 52E

a.

To determine

To compare the distance between tree at a point $A$ with point $B$ and point $C$ .

Answer to Problem 52E

The tree at point $B$ is farther compare to the tree at point $C$ from the tree at a point $A$ is $82-78=4\text{ meters}$

Explanation of Solution

Given: $1$ unit = $10$ meters.

  

Calculation:

  $\begin{array}{l}A=(10,10)=(x_1,y_1)\\ C=(60,70)=(x_2,y_2)\\ The\text{ distance between A and C is calculated using the formula:}\\ d=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}=\sqrt{{(60-10)}^2+{(70-10)}^2}\\ =\sqrt{{(50)}^2+{(60)}^2}=\sqrt{2500+3600}=\sqrt{6100}=78.1\approx 78meters.\\ And\\ A=(10,10)=(x_1,y_1)\\ B=(90,30)=(x_2,y_2)\\ The\text{ distance between A and B is calculated using the formula:}\\ d=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}=\sqrt{{(90-10)}^2+{(30-10)}^2}\\ =\sqrt{{(80)}^2+{(20)}^2}=\sqrt{6400+400}=\sqrt{6800}=82.46\approx \text{82 }meters.\end{array}$

Hence, the tree at point $B$ is farther compare to the tree at point $C$ from the tree at a point $A$ = $82-78=4\text{ meters}$

b.

To determine

To find the distance of new tree.

Answer to Problem 52E

The new tree is at a distance of $39$ meters.

Explanation of Solution

Given:

  

  $\begin{array}{l}A=(10,10)=(x_1,y_1)\\ B=(90,30)=(x_2,y_2)\\ The{\text{ centroid or midpoint between A and B =(X}}_1{\text{,Y}}_1\text{)=}\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)=\left(\frac{10+90}{2},\frac{10+30}{2}\right)=\left(\frac{100}{2},\frac{40}{2}\right)=\left(50,20\right)\\ \text{ Similarly, }\\ B=(90,30)=(x_1,y_1)\\ C=(60,70)=(x_2,y_2)\\ The{\text{ centroid or midpoint between B and C =(X}}_2{\text{,Y}}_2\text{)=}\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)=\left(\frac{90+60}{2},\frac{30+70}{2}\right)=\left(\frac{150}{2},\frac{100}{2}\right)=\left(75,50\right)\\ \text{Then the distance between new trees = D =}\sqrt{\left(X_2-X_1{)}^{{}^2}\right)+{\left(Y_2-Y_1\right)}^{{}^2}}=\sqrt{\left(75-50{)}^{{}^2}\right)+{\left(50-20\right)}^{{}^2}}\\ \text{ D}=\sqrt{{25}^2+{30}^2}=\sqrt{625+900}=\sqrt{1525}=39meters\\ \text{Hence the new tree is at distance of 39 meters}\text{.}\end{array}$