Glencoe Math: Course 3, Volume 2
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Glencoe Math: Course 3, Volume 2 View details
7. Distance on the Coordinate Plane
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Exercise 9 Page 436

Practice makes perfect

Let's start by recalling that an ( x, y) ordered pair is plotted in a coordinate plane by finding the x-coordinate on the horizontal axis and the y-coordinate on the vertical axis. For point A( - 2, 1), this means moving two units in the negative horizontal direction starting from the origin, then one unit in the positive vertical direction.

The coordinate plane

To graph point B( - 2, 6), we need to move two units in the negative horizontal direction starting from the origin, then six units in the positive vertical direction.

The coordinate plane

Finally, let's graph point C( 1, 3) by moving one unit in the positive horizontal direction starting from the origin and three units in the positive vertical direction.

The coordinate plane
We can use the Distance Formula to find the length of segment BC.

d=sqrt(( x_2- x_1)^2+( y_2- y_1)^2) In the formula, d is the distance between two points with coordinates ( x_1, y_1) and ( x_2, y_2). Since the length of a segment is the distance between two points, we can find it using this formula. To do so, we will substitute the coordinates of the points B and C into the formula and simplify.

Let's start by connecting the points that we graphed in Part A.

The triangle
To find the length of each side of △ ABC, we will use the Distance Formula. d=sqrt(( x_2- x_1)^2+( y_2- y_1)^2) In the formula, d is the distance between two points with coordinates ( x_1, y_1) and ( x_2, y_2). Since each side length is a distance between two points, we can find it using this formula. First, we will find the distance between the points A( - 2, 1) and C( 1, 3). To do so, we will substitute the given coordinates into the formula and simplify.
d = sqrt((x_2-x_1)^2 + (y_2-y_1)^2)
d=sqrt(( 1-( - 2))^2+( 3- 1)^2)
d=sqrt((1+2)^2+(3-1)^2)
d=sqrt(3^2+2^2)
d=sqrt(9+4)
d=sqrt(13)
d=3.605551 ...
d≈ 3.6
Therefore, the length of segment AC is equal to about 3.6 units. Next, we will find the distance between the points A( - 2, 1) and B( - 2, 6).
d = sqrt((x_2-x_1)^2 + (y_2-y_1)^2)
d=sqrt(( - 2-( - 2))^2+( 6- 1)^2)
d=sqrt((- 2+2)^2+(6-1)^2)
d=sqrt(0^2+5^2)
d=sqrt(0+25)
d=sqrt(25)
d=5
We got that the length of segment AB is equal to 5 units. Finally, we will find the distance between the points B( - 2, 6) and C( 1, 3).
d = sqrt((x_2-x_1)^2 + (y_2-y_1)^2)
d=sqrt(( 1-( - 2))^2+( 3- 6)^2)
d=sqrt((1+2)^2+(3-6)^2)
d=sqrt(3^2+(- 3)^2)
d=sqrt(3^2+3^2)
d=sqrt(9+9)
d=sqrt(18)
d = 4.242640 ...
d ≈ 4.2
Therefore, the length of segment BC is about 4.2 units.
To determine the perimeter of triangle ABC, all of its side lengths must be added. In Part C, we found that AB ≈ 3.6, AB=5, and BC ≈ 4.2. Let's add these side lengths together!

3.6+5+4.2=12.8 We got that the perimeter of triangle ABC is equal to about 12.8 units.