Q16. - Algebra 1

Question

Karen is making a map of her hometown using a coordinate grid. The scale of the map is $\mathbf{1}\mathit{u}\mathit{n}\mathit{i}\mathit{t}\mathbf{=}\mathbf{2}\mathbf{.5}\mathit{m}\mathit{i}\mathit{l}\mathit{e}\mathit{s}$.

a. What is the actual distance between Karen’s school and the park? Round to the nearest tenth of a mile if necessary.

 

b. Suppose Karen’s house is located midway between the mall and the school. What coordinates represent her house? Show the work.

Step-by-Step Solution

Answer

a. The distance between Karen’s school and the park is$1unit=2.5miles$ $\mathbf{25}\mathbf{.1}\mathit{m}\mathit{i}\mathit{l}\mathit{e}\mathit{s}$.

 

b. The coordinates of Karen’s House is $\left(\frac{{\displaystyle 1}}{{\displaystyle 2}},\frac{{\displaystyle 1}}{{\displaystyle 2}}\right)$.

Part a Step 1. State the concept of ‘Distance between two points in 2-dimensional plane’.

Let the two points in the 2-dimensional plane be $(x_1,y_1)$ and $(x_2,y_2)$.

The distance between the points $P$ and $Q$ is given as:

$d=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}\dots \left(1\right)$

Part a Step 2. State the concept of ‘rounding to nearest tenth’.

Suppose there is more than one digit after decimal then we round up to the $1^{st}$ decimal number which is called the tenth digit using the following rules. The number after the tenth digit is called the hundredth digit.

If the hundredth digit is less than 5, then keep the tenth digit unchanged and rewrite the number by removing decimal digits after tenths.

If the hundredth digit is greater than or equal to 5, then add 1 to the tenth digit and rewrite the number by removing decimal digits after tenths.

For example:

$6.57$ is rounded to $6.6$. As the hundredth digit is 7, which is greater than 5.

$6.53$ is rounded to $6.5$. The tenths digit 5 is kept unchanged as the hundredths digit 3 is less than 5.

Part a Step 3. Calculate the distance between Karen’s school and the park.

Observe the figure given below.

From the figure, 

The coordinate of Karen’s school is  and the coordinates of the park are .

Substitute $(4,5)$ as $(x_1,y_1)$ and $(5,-5)$ as $(x_2,y_2)$ in equation $\left(1\right)$ to find the distance.

$\begin{array}{l}d=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}\\ =\sqrt{{(5-4)}^2+{(-5-5)}^2}\\ =\sqrt{{\left(1\right)}^2+{(-10)}^2}\\ =\sqrt{1+100}\\ =\sqrt{101}\\ =10.05\end{array}$

The scale of the map is $1unit=2.5miles$.

$\begin{array}{l}\text{Therefore for 10.05 units}=10.05\times 2.5\\ =25.1 \left[\text{Rounded\hspace{0.17em}\hspace{0.17em}to\hspace{0.17em}\hspace{0.17em}nearest\hspace{0.17em}\hspace{0.17em}tenth}\right]\end{array}$

Hence, the distance between Karen’s school and the park is approximately $25.1$ miles.

Part b Step 1. State the concept of ‘midpoint of a segment in 2-dimensional plane’.

Let the endpoints of the segment in a 2-dimensional plane be $(x_1,y_1)$ and $(x_2,y_2)$.

The midpoint of the segment with endpoints is given as:

$(x,y)=\left(\frac{{\displaystyle x_1+x_2}}{{\displaystyle 2}},\frac{{\displaystyle y_1+y_2}}{{\displaystyle 2}}\right)\dots \left(2\right)$

Here $(x,y)$ is the midpoint.

Part b Step 2. Find the coordinates of Karen’s house.

Observe the figure given below.

From the figure, 

The coordinate of the mall is $(-3,-4)$ and the coordinate of the school is $(4,5)$.

Karen’s house is located midway between the mall and the school.

Substitute $(-3,-4)$ as $(x_1,y_1)$ and $(4,5)$ as $(x_2,y_2)$ in equation $\left(2\right)$ to find the coordinate of Karen’s house.

$\begin{array}{c}(x,y)=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)\\ =\left(\frac{-3+4}{2},\frac{-4+5}{2}\right)\\ =\left(\frac{1}{2},\frac{1}{2}\right)\end{array}$

Part b Step 3. State the conclusion.

Therefore, the coordinates of Karen’s house are $\left(\frac{{\displaystyle 1}}{{\displaystyle 2}},\frac{{\displaystyle 1}}{{\displaystyle 2}}\right)$.