Question

How do I find the difference between two points on a coordinate plane?
On a coordinate plane, Point P is located at (2,-3) and Point Q is located at (-4,4). Find the difference between Point P and Point Q. Round your answer to the nearest tenth.
Explain or show all your work.

Answer 1

https://www.khanacademy.org/math/cc-sixth-grade-math/cc-6th-negative-number-topic/cc-6th-coordinate-plane/v/the-coordinate-plane see if this helps

Answer 2

that link wouldnt really help would it?

we actually have to use a formula which looks like this:
\[ \large\text{D} \scriptsize\text{istance}= \sqrt{ (\color{coral}{x_2} -\color{lightskyblue}{x_1})^2 + (\color{gold}{y_2} - \color{lightseagreen}{y_1})^2 } \]

where

\[ \color{coral}{x_2} = 2\]

\[\color{lightskyblue}{x_1} = -4\]

\[\color{gold}{y_2} = -3\]

\[\color{lightseagreen}{y_1} = 4\]
thus, by substituting the numbers in we get:
\[ \large\text{D} \scriptsize\text{istance}= \sqrt{ (\color{coral}{2} -(\color{lightskyblue}{-4}))^2 + (\color{gold}{-3} - \color{lightseagreen}{4})^2 } \]
\[ \large\text{D} \scriptsize\text{istance}= \sqrt{ (\color{coral}{2} +\color{lightskyblue}{4})^2 + (\color{gold}{-3} - \color{lightseagreen}{4})^2 } \]
\[ \large\text{D} \scriptsize\text{istance}= \sqrt{ 6^2 + -7^2 } \]
\[ \large\text{D} \scriptsize\text{istance}= \sqrt{ 36 + 49 } \]
\[ \large\text{D} \scriptsize\text{istance}= \sqrt{ 85 } \]
\[ \large\text{D} \scriptsize\text{istance}= ?????~~~\text{ ( round it}) \]

Answer 3

start by usg the point of orgin and gon negative or positivly up or down