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 24 Page 438

The x-coordinate represents the horizontal shift, starting from the origin, and the y-coordinate represents the vertical shift.

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 X( - 1, - 2), this means moving one unit in the negative horizontal direction starting from the origin, then two units in the negative vertical direction.

The coordinate plane

To graph point Y( 1, 2), we need to move one unit in the positive horizontal direction starting from the origin, then two units in the positive vertical direction.

The coordinate plane

Finally, we will graph point Z( 3, -2) by moving three units in the positive horizontal direction starting from the origin, then two units in the negative vertical direction.

The coordinate plane

Now we can connect the points by drawing three segments XY, YZ, and ZX.

The coordinate plane
We got a triangle. To determine whether the triangle is equilateral, isosceles, or scalene, we can calculate all the side lengths of the triangle. Let's do this using 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 the distance between two points, we can find it using this formula. First, let's find the distance between the points X( - 1, - 2) and Y( 1, 2) by substituting the given coordinates into the formula and simplifying.
d = sqrt((x_2-x_1)^2 + (y_2-y_1)^2)
|XY|=sqrt(( 1-( - 1))^2+( 2-( - 2))^2)
|XY|=sqrt((1+1)^2+(2+2)^2)
|XY|=sqrt(2^2+4^2)
|XY|=sqrt(4+16)
|XY|=sqrt(20)
Next, we will find the distance between the points Y( 1, 2) and Z( 3, - 2).
d = sqrt((x_2-x_1)^2 + (y_2-y_1)^2)
|YZ|=sqrt(( 3- 1)^2+( - 2- 2)^2)
|YZ|=sqrt(2^2+(- 4)^2)
|YZ|=sqrt(2^2+4^2)
|YZ|=sqrt(4+16)
|YZ|=sqrt(20)
Finally, we will find the distance between the points Z( 3, - 2) and X( - 1, - 2).
d = sqrt((x_2-x_1)^2 + (y_2-y_1)^2)
|ZX|=sqrt(( - 1- 3)^2+( - 2-( - 2))^2)
|ZX|=sqrt((- 1-3)^2+(- 2+2)^2)
|ZX|=sqrt((- 4)^2+0^2)
|ZX|=sqrt(4^2+0^2)
|ZX|=sqrt(16+0)
|ZX|=sqrt(16)
|ZX|=4
Therefore, we got that the side lengths of the triangle are sqrt(20), sqrt(20), and 4. Since two sides of the triangle have equal lengths, the triangle is isosceles.