Question

find the distance of the two given points:

1. (-1, -3), (8, 6)
2. (-1, 3), (7, -5)
3. (-7, 1), (2, 8)
4. (-2, -1, (5, -4)
5. (-5, 8), (-9, -6)
6. (7, 6), (2, -9)

Explanation:

Step1: Recall 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 the first pair of points $(-1,-3)$ and $(8,6)$

$x_1=-1,y_1 = - 3,x_2=8,y_2 = 6$
$d_1=\sqrt{(8-(-1))^2+(6 - (-3))^2}=\sqrt{(8 + 1)^2+(6 + 3)^2}=\sqrt{9^2+9^2}=\sqrt{81 + 81}=\sqrt{162}=9\sqrt{2}$

Step3: Solve for the second pair of points $(-1,3)$ and $(7,-5)$

$x_1=-1,y_1 = 3,x_2=7,y_2=-5$
$d_2=\sqrt{(7-(-1))^2+(-5 - 3)^2}=\sqrt{(7 + 1)^2+(-8)^2}=\sqrt{8^2+64}=\sqrt{64 + 64}=\sqrt{128}=8\sqrt{2}$

Step4: Solve for the third pair of points $(-7,1)$ and $(2,8)$

$x_1=-7,y_1 = 1,x_2=2,y_2=8$
$d_3=\sqrt{(2-(-7))^2+(8 - 1)^2}=\sqrt{(2 + 7)^2+7^2}=\sqrt{9^2+49}=\sqrt{81+49}=\sqrt{130}$

Step5: Solve for the fourth pair of points $(-2,-1)$ and $(5,-4)$

$x_1=-2,y_1=-1,x_2=5,y_2=-4$
$d_4=\sqrt{(5-(-2))^2+(-4-(-1))^2}=\sqrt{(5 + 2)^2+(-3)^2}=\sqrt{49 + 9}=\sqrt{58}$

Step6: Solve for the fifth pair of points $(-5,8)$ and $(-9,-6)$

$x_1=-5,y_1 = 8,x_2=-9,y_2=-6$
$d_5=\sqrt{(-9-(-5))^2+(-6 - 8)^2}=\sqrt{(-4)^2+(-14)^2}=\sqrt{16+196}=\sqrt{212}=2\sqrt{53}$

Step7: Solve for the sixth pair of points $(7,6)$ and $(2,-9)$

$x_1=7,y_1 = 6,x_2=2,y_2=-9$
$d_6=\sqrt{(2 - 7)^2+(-9 - 6)^2}=\sqrt{(-5)^2+(-15)^2}=\sqrt{25+225}=\sqrt{250}=5\sqrt{10}$

Answer:

1. $9\sqrt{2}$
2. $8\sqrt{2}$
3. $\sqrt{130}$
4. $\sqrt{58}$
5. $2\sqrt{53}$
6. $5\sqrt{10}$