Use the coordinate plane to estimate the distance between the two points

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Use the coordinate plane to estimate the distance between the two points. Then use the distance formula to find the distance between the points. Round the result to the nearest hundredth If necessary. |(4, 3) 3. 1. 2. y -3. 3- (2, 3) (-3, 1) 5 x -3 3 x (1, – 4) (-3, -3) (2, –3) Find the distance between the two points. Round the result to the nearest hundredth if necessary. 4. (1, 1), (4, 4) 5. (2, 5), (5, 1) 6. (0, 3), (2, 6) 9. (3, -5), (–2, 0) 7. (1, 6), (5, 1) 10. (-3,-5), (6, 5) 8. (-2, 8), (4, 0) 11. (8, 6), (→4, – 3) 12. (-5, 2), (–2, 5) Lesson 12,6 None

The correct answer and explanation is:

To find the distance between two points on a coordinate plane, you can use the distance formula, which is derived from the Pythagorean theorem. The distance formula is: d=(x2−x1)2+(y2−y1)2d = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}

Where:

  • (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) are the coordinates of the two points.

Example Calculations:

1. Distance between (1, 1) and (4, 4):

  • x1=1x_1 = 1, y1=1y_1 = 1, x2=4x_2 = 4, y2=4y_2 = 4
  • Using the distance formula:

d=(4−1)2+(4−1)2=32+32=9+9=18≈4.24d = \sqrt{(4 – 1)^2 + (4 – 1)^2} = \sqrt{3^2 + 3^2} = \sqrt{9 + 9} = \sqrt{18} \approx 4.24

So, the distance is approximately 4.24 units.

2. Distance between (2, 5) and (5, 1):

  • x1=2x_1 = 2, y1=5y_1 = 5, x2=5x_2 = 5, y2=1y_2 = 1
  • Using the distance formula:

d=(5−2)2+(1−5)2=32+(−4)2=9+16=25=5d = \sqrt{(5 – 2)^2 + (1 – 5)^2} = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5

So, the distance is 5 units.

3. Distance between (0, 3) and (2, 6):

  • x1=0x_1 = 0, y1=3y_1 = 3, x2=2x_2 = 2, y2=6y_2 = 6
  • Using the distance formula:

d=(2−0)2+(6−3)2=22+32=4+9=13≈3.61d = \sqrt{(2 – 0)^2 + (6 – 3)^2} = \sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13} \approx 3.61

So, the distance is approximately 3.61 units.

4. Distance between (3, -5) and (-2, 0):

  • x1=3x_1 = 3, y1=−5y_1 = -5, x2=−2x_2 = -2, y2=0y_2 = 0
  • Using the distance formula:

d=(−2−3)2+(0−(−5))2=(−5)2+52=25+25=50≈7.07d = \sqrt{(-2 – 3)^2 + (0 – (-5))^2} = \sqrt{(-5)^2 + 5^2} = \sqrt{25 + 25} = \sqrt{50} \approx 7.07

So, the distance is approximately 7.07 units.

Explanation:

The distance formula works by first calculating the horizontal and vertical differences between the two points. These differences form the legs of a right triangle, where the distance between the points is the hypotenuse. By applying the Pythagorean theorem, we compute the square of these differences, sum them, and take the square root to get the distance.

Each of the distances above is rounded to the nearest hundredth if necessary. For most of these problems, it’s important to use the correct formula and round only at the end to avoid errors.

Feel free to ask for help with more points!

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