Question

find the distance of the two given points:
(-1, -3), (8, 6)
(-1, 3), (7, -5)
(-7, 1), (2, 8)
(-2, -1), (5, -4)
(-5, 8), (-9, -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: For 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: For 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: For 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: For points $(-2,-1)$ and $(5,-4)$

$x_1=-2,y_1=-1,x_2=5,y_2=-4$.
$d_4=\sqrt{(5-(-2))^2+(-4 - (-1))^2}=\sqrt{(5 + 2)^2+(-4 + 1)^2}=\sqrt{7^2+(-3)^2}=\sqrt{49 + 9}=\sqrt{58}$.

Step6: For points $(-5,8)$ and $(-9,-6)$

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

Step7: For 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:

The distances are: $9\sqrt{2},8\sqrt{2},\sqrt{130},\sqrt{58},2\sqrt{53},5\sqrt{10}$ for the respective pairs of points.