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# The Cartesian Plane Help (page 2)

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## Distance Between Points

Suppose there are two different points P = ( x 0 , y 0 ) and Q = ( x 1 , y 1 ) on the Cartesian plane. The distance d between these two points can be found by determining the length of the hypotenuse, or longest side, of a right triangle PQR , where point R is the intersection of a “horizontal” line through P and a “vertical” line through Q . In this case, “horizontal” means “parallel to the x axis,” and “vertical” means “parallel to the y axis.” An example is shown in Fig. 6-2. Alternatively, we can use a “horizontal” line through Q and a “vertical” line through P to get the point R . The resulting right triangle in this case has the same hypotenuse, line segment PQ , as the triangle determined the other way.

Fig. 6-2 . Two points P and Q , plotted in rectangular coordinates, and a third point R , important in finding the distance d between P and Q.

Recall the Pythagorean theorem. It states that the square of the length of the hypotenuse of a right triangle is equal to the sum of the squares of the other two sides. In this case, that means:

d 2 = ( x 1x 0 ) 2 + ( y 1y 0 ) 2

and therefore:

d = [( x 1x 0 ) 2 + ( y 1y 0 ) 2 ] 1/2

where the 1/2 power is the square root. In the situation shown in Fig. 6-2, the distance d between points P = ( x 0 , y 0 ) = (−5, −4) and Q = ( x 1 , y 1 ) = (3,5) is:

d ={[3 −(−5)] 2 + [5−(−4)] 2 } 1/2

= [(3 + 5) 2 + (5 + 4) 2 ] 1/2

= (8 2 + 9 2 ) 1/2

= (64 + 81) 1/2

= 145 1/2

= 12.04(approx.)

This is accurate to two decimal places, as determined using a standard digital calculator that can find square roots.

## The Cartesian Plane Practice Problems

#### PROBLEM 1

Plot the following points on the Cartesian plane: (−2,3), (3,−1), (0,5), and (−3,−3).

#### SOLUTION 1

These points are shown in Fig. 6-4. The dashed lines are perpendiculars, dropped to the axes to show the x and y coordinates of each point. (The dashed lines are not parts of the coordinates themselves.)

Fig. 6-4 . Illustration for Problems 1 and 2.

#### PROBLEM 2

What is the distance between (0,5) and (−3, −3) in Fig. 6-4? Express the answer to three decimal places.

#### SOLUTION 2

Use the distance formula. Let ( x 0 , y 0 ) = (0,5) and ( x 1 , y 1 ) = (−3, −3). Then:

d = [( x 1x 0 ) 2 + ( y 1y 0 ) 2 ] 1/2

= [(−3 −0) 2 + (−3 − 5) 2 ] 1/2

= [(−3) 2 + (−8) 2 ] 1/2

= (9 + 64) 1/2

= 73 1/2

= 8.544 (approx.)

Practice problems for these concepts can be found at: The Cartesian Plane Practice Test.

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