Question

topic 1 using a rectangular coordinate system skills practice continued
3 a (5, 8), b (5, 2), c (-3, 5)
4 a (-2, -6), b (6, -6), c (2, -3)
ab $sqrt{(5 - 5)^2+(2 - 8)^2}=6$
bc $sqrt{(-3 - 5)^2+(5 - 2)^2}=\sqrt{73}$
ca $sqrt{(-3 - 5)^2+(5 - 8)^2}=\sqrt{73}$
5 a (0, 0), b (4, 0), c (3, 7)
6 a (-6, 4), b (0, 4), c (-2, -2)

Explanation:

Step1: Recall distance formula

The distance $d$ 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: Calculate distance for problem 3 - AB

For points $A(5,8)$ and $B(5,2)$, substitute into the formula: $AB=\sqrt{(5 - 5)^2+(2 - 8)^2}=\sqrt{0+( - 6)^2}=\sqrt{36}=6$.

Step3: Calculate distance for problem 3 - BC

For points $B(5,2)$ and $C(-3,5)$, $BC=\sqrt{(-3 - 5)^2+(5 - 2)^2}=\sqrt{(-8)^2+3^2}=\sqrt{64 + 9}=\sqrt{73}$.

Step4: Calculate distance for problem 3 - CA

For points $C(-3,5)$ and $A(5,8)$, $CA=\sqrt{(5+3)^2+(8 - 5)^2}=\sqrt{8^2+3^2}=\sqrt{64+9}=\sqrt{73}$.

Step5: Calculate distance for problem 5 - AB

For points $A(0,0)$ and $B(4,0)$, $AB=\sqrt{(4 - 0)^2+(0 - 0)^2}=\sqrt{4^2+0}=\sqrt{16}=4$.

Step6: Calculate distance for problem 5 - BC

For points $B(4,0)$ and $C(3,7)$, $BC=\sqrt{(3 - 4)^2+(7 - 0)^2}=\sqrt{(-1)^2+7^2}=\sqrt{1 + 49}=\sqrt{50}=5\sqrt{2}$.

Step7: Calculate distance for problem 5 - CA

For points $C(3,7)$ and $A(0,0)$, $CA=\sqrt{(0 - 3)^2+(0 - 7)^2}=\sqrt{(-3)^2+( - 7)^2}=\sqrt{9 + 49}=\sqrt{58}$.

Step8: Calculate distance for problem 6 - AB

For points $A(-6,4)$ and $B(0,4)$, $AB=\sqrt{(0 + 6)^2+(4 - 4)^2}=\sqrt{6^2+0}=\sqrt{36}=6$.

Step9: Calculate distance for problem 6 - BC

For points $B(0,4)$ and $C(-2,-2)$, $BC=\sqrt{(-2 - 0)^2+(-2 - 4)^2}=\sqrt{(-2)^2+( - 6)^2}=\sqrt{4 + 36}=\sqrt{40}=2\sqrt{10}$.

Step10: Calculate distance for problem 6 - CA

For points $C(-2,-2)$ and $A(-6,4)$, $CA=\sqrt{(-6 + 2)^2+(4 + 2)^2}=\sqrt{(-4)^2+6^2}=\sqrt{16+36}=\sqrt{52}=2\sqrt{13}$.

Answer:

For problem 3: $AB = 6$, $BC=\sqrt{73}$, $CA=\sqrt{73}$; For problem 5: $AB = 4$, $BC=5\sqrt{2}$, $CA=\sqrt{58}$; For problem 6: $AB = 6$, $BC=2\sqrt{10}$, $CA=2\sqrt{13}$