Q1. - Geometry

Question

Copy the grid system shown on the previous page onto a piece of graph paper. Then locate the following points.

a. A point $\mathit{T}$ five blocks due west of the centre of town.

b. A point $\mathit{U}$ five blocks east and two blocks south of the centre of town

c. A point $\mathit{V}$ two blocks west and one block north of your house, which is located at point $\mathit{P}$

Step-by-Step Solution

Answer

Points $T$, $U$ and $V$ are located in the following graph

Part a. Step 1 - Directions on grid system

If centre of the town is at the origin then 

• east side is towards positive $x\text{-axis}$
• west side is towards negative $x\text{-axis}$
• north side is towards positive $y\text{-axis}$
• south side is towards negative $y\text{-axis}$

Part a. Step 2 - Find coordinate of point T

As the point is 5 units towards negative $x\text{-axis}$, so its coordinate is $\left(-5,0\right)$

Part a. Step 3 - Plot the point

To show point $T\left(-5,0\right)$ on coordinate grid mark the dot as shown below

Part b. Step 1 - Directions on grid system

If centre of the town is at the origin then 

• east side is towards positive $x\text{-axis}$
• west side is towards negative $x\text{-axis}$
• north side is towards positive $y\text{-axis}$
• south side is towards negative $y\text{-axis}$

Part b. Step 2 - Find coordinate of point U

As the point is 5 units towards positive $x\text{-axis}$ and 2 units towards negative $y\text{-axis}$, so its coordinate is $\left(5,-2\right)$

Part b. Step 3 - Plot the point

To show point $U\left(5,-2\right)$ on coordinate grid mark the dot as shown below

Part c. Step 1 - Directions on grid system

If centre of the town is at the origin then

• east side is towards positive $x\text{-axis}$
• west side is towards negative $x\text{-axis}$
• north side is towards positive $y\text{-axis}$
• south side is towards negative $y\text{-axis}$

Part c. Step 2 - Find coordinate of point V

As the point is 2 units towards negative $x\text{-axis}$ and 1 unit towards positive $y\text{-axis}$ from point $P$, so subtract 2 from $x\text{-coordinate}$ and add 1 in $y\text{-coordinate}$ of point $P$ to get,

$\begin{array}{rcl}\left(3,2\right)+\left(-2,1\right)& =& \left(3-2,2+1\right)\\ & =& \left(1,3\right)\end{array}$

Part c. Step 3 - Plot the point

To show point $V\left(1,3\right)$ on coordinate grid mark the dot as shown below