.
|dw:1441573642078:dw|
@freckles
n this case, you need the distance formula.
Those are ordered pairs, right? That show where each point is located?
ya
Do you know the distance formula?
sqrt of x2-x1 ^2+ y2-y2^2 pronounced y-1, not y ^2
Yes. So in order to find the area, you have to find the distance between the length and the width seperately. First, do (-5, 3), and (7, -6).
Just substitute and solve. Let me know if you need help.
64+169
want me to continue or something???
Yes. I will check your answers as you go, if you would like.
233 now what???
You have to do the same thing with the points (-2, 7) and (-5, 3)
81+64 145
In order to find the width. Then, you just multiply the 2. I will be checking your work,
what do i do in order to find the width??
To find the width, use the distance formula for these 2 points: (-2, 7) and (-5, 3)
yah wchich is 145
Yes. To find the area, just multiply the length and width together.
um that's 33785 AAAANNNNDDD thats not an answer choice you are wrong
I am checking your work, and a step was done wrong.
For the width, it is 5.
And I am double checking my work for the length,
how do you get 5??????
Ok so, Sqrt (-5 + 2)^2 + (3 - 7)^2 --> Sqrt (-3)^2 + (-4)^2 --> Sqrt (9 + 16) --> 5
okay okaywhats the lenght
15
But do you understand the concept?
"Yes. So in order to find the area, you have to find the distance between the length and the width seperately. First, do (-5, 3), and (7, -6). " \[\text{ distance between these two points is } \\ \sqrt{(-5-7)^2+(3-(-6))^2} \\ =\sqrt{(-5-7)^2+(3+6)^2} \\ =\sqrt{(-12)^2+(9)^2} .. \text{ so on ... }\]
yeah so 75
Yes. 75 is correct.
can you help with one more?
I will try, but I am not very good at geometry, as I just started learning it. But shoot.
|dw:1441574726192:dw| find the perimeter
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