Question

software - infinite algebra 1
the distance formula
find the distance between each pair of p

1. (7, 3), (-1, -4)
2. (6, -7), (3, -5)
3. (5, -8), (-8, 6)
4. (-7, 0), (-2, -4)
5. (-2, 2), (-6, -8)
6. (-3, -1), (-4, 0)
7. (-2, 3), (-1, 7)

Explanation:

Step1: Recall the Distance Formula

The distance formula between two points \((x_1, y_1)\) and \((x_2, y_2)\) is \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\).

Step2: Solve for Problem 1: \((7, 3)\) and \((-1, -4)\)

Step 2.1: Identify \(x_1, y_1, x_2, y_2\)

Here, \(x_1 = 7\), \(y_1 = 3\), \(x_2 = -1\), \(y_2 = -4\).

Step 2.2: Substitute into the formula

\(d=\sqrt{(-1 - 7)^2+(-4 - 3)^2}=\sqrt{(-8)^2+(-7)^2}\)

Step 2.3: Calculate the squares

\((-8)^2 = 64\) and \((-7)^2=49\)

Step 2.4: Sum the squares and take the square root

\(d=\sqrt{64 + 49}=\sqrt{113}\) (Note: The original answer had a mistake, the correct distance is \(\sqrt{113}\approx10.63\), not \(113\))

Step3: Solve for Problem 3: \((6, -7)\) and \((3, -5)\)

Step 3.1: Identify \(x_1, y_1, x_2, y_2\)

\(x_1 = 6\), \(y_1=-7\), \(x_2 = 3\), \(y_2=-5\)

Step 3.2: Substitute into the formula

\(d=\sqrt{(3 - 6)^2+(-5-(-7))^2}=\sqrt{(-3)^2+(2)^2}\)

Step 3.3: Calculate the squares

\((-3)^2 = 9\) and \(2^2 = 4\)

Step 3.4: Sum the squares and take the square root

\(d=\sqrt{9 + 4}=\sqrt{13}\) (This matches the original answer)

Step4: Solve for Problem 5: \((5, -8)\) and \((-8, 6)\)

Step 4.1: Identify \(x_1, y_1, x_2, y_2\)

\(x_1 = 5\), \(y_1=-8\), \(x_2=-8\), \(y_2 = 6\)

Step 4.2: Substitute into the formula

\(d=\sqrt{(-8 - 5)^2+(6-(-8))^2}=\sqrt{(-13)^2+(14)^2}\)

Step 4.3: Calculate the squares

\((-13)^2=169\) and \(14^2 = 196\)

Step 4.4: Sum the squares and take the square root

\(d=\sqrt{169+196}=\sqrt{365}\approx19.10\) (The original had a typo with "16a", correct is 169)

Step5: Solve for Problem 7: \((-7, 0)\) and \((-2, -4)\)

Step 5.1: Identify \(x_1, y_1, x_2, y_2\)

\(x_1=-7\), \(y_1 = 0\), \(x_2=-2\), \(y_2=-4\)

Step 5.2: Substitute into the formula

\(d=\sqrt{(-2-(-7))^2+(-4 - 0)^2}=\sqrt{(5)^2+(-4)^2}\)

Step 5.3: Calculate the squares

\(5^2 = 25\) and \((-4)^2=16\)

Step 5.4: Sum the squares and take the square root

\(d=\sqrt{25 + 16}=\sqrt{41}\approx6.40\) (Original had incorrect steps, this is the correct calculation)

Step6: Solve for Problem 9: \((-2, 2)\) and \((-6, -8)\)

Step 6.1: Identify \(x_1, y_1, x_2, y_2\)

\(x_1=-2\), \(y_1 = 2\), \(x_2=-6\), \(y_2=-8\)

Step 6.2: Substitute into the formula

\(d=\sqrt{(-6-(-2))^2+(-8 - 2)^2}=\sqrt{(-4)^2+(-10)^2}\)

Step 6.3: Calculate the squares

\((-4)^2 = 16\) and \((-10)^2=100\)

Step 6.4: Sum the squares and take the square root

\(d=\sqrt{16 + 100}=\sqrt{116}=2\sqrt{29}\approx10.77\)

Step7: Solve for Problem 11: \((-3, -1)\) and \((-4, 0)\)

Step 7.1: Identify \(x_1, y_1, x_2, y_2\)

\(x_1=-3\), \(y_1=-1\), \(x_2=-4\), \(y_2 = 0\)

Step 7.2: Substitute into the formula

\(d=\sqrt{(-4-(-3))^2+(0-(-1))^2}=\sqrt{(-1)^2+(1)^2}\)

Step 7.3: Calculate the squares

\((-1)^2 = 1\) and \(1^2 = 1\)

Step 7.4: Sum the squares and take the square root

\(d=\sqrt{1 + 1}=\sqrt{2}\approx1.41\)

Step8: Solve for Problem 13: \((-2, 3)\) and \((-1, 7)\)

Step 8.1: Identify \(x_1, y_1, x_2, y_2\)

\(x_1=-2\), \(y_1 = 3\), \(x_2=-1\), \(y_2 = 7\)

Step 8.2: Substitute into the formula

\(d=\sqrt{(-1-(-2))^2+(7 - 3)^2}=\sqrt{(1)^2+(4)^2}\)

Step 8.3: Calculate the squares

\(1^2 = 1\) and \(4^2=16\)

Step 8.4: Sum the squares and take the square root

\(d=\sqrt{1 + 16}=\sqrt{17}\approx4.12\)

Answer:

1. \(\boldsymbol{\sqrt{113}}\) (≈10.63)
2. \(\boldsymbol{\sqrt{13}}\) (≈3.61)
3. \(\boldsymbol{\sqrt{365}}\) (≈19.10)
4. \(\boldsymbol{\sqrt{41}}\) (≈6.40)
5. \(\boldsymbol{2\sqrt{29}}\) (≈10.77)
6. \(\boldsymbol{\sqrt{2}}\) (≈1.41)
7. \(\boldsymbol{\sqrt{17}}\) (≈4.12)