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問題

解法ステップ

1. Understand the question

   Turn 1 Question

   finding distance on the coordinate plane

   1. plot the points
   2. create a triangle
   3. find the distance between the points

   point a (-4, 1) point b (2, 4)
   $|-4-2|=6$
   $|1-4|=3$
   $a^2+b^2=c^2$
   $6^2+3^2=c^2$
   $36+9=45$
   $sqrt{45}=6.7$
   point c (-5, 2) point d (-2, -4)
   point e (3,1) point f (2, -3)
   point g (2,-2) point h (3, -4)

   Turn 2 Question

   Can u turn them to a image

2. Response

   Turn 1 Answer

3. Explanation

   Step1: Correct A-B distance calculation

   First, confirm horizontal/vertical differences:
   $\Delta x = |-4 - 2| = 6$, $\Delta y = |1 - 4| = 3$
   Apply distance formula: $d=\sqrt{(\Delta x)^2+(\Delta y)^2}=\sqrt{6^2+3^2}=\sqrt{36+9}=\sqrt{45}=3\sqrt{5}\approx6.7$

   Step2: Calculate C-D distance

   Find $\Delta x = |-5 - (-2)| = 3$, $\Delta y = |2 - (-4)| = 6$
   $d_{CD}=\sqrt{3^2+6^2}=\sqrt{9+36}=\sqrt{45}=3\sqrt{5}\approx6.7$

   Step3: Calculate E-F distance

   Find $\Delta x = |3 - 2| = 1$, $\Delta y = |1 - (-3)| = 4$
   $d_{EF}=\sqrt{1^2+4^2}=\sqrt{1+16}=\sqrt{17}\approx4.1$

   Step4: Calculate G-H distance

   Find $\Delta x = |2 - 3| = 1$, $\Delta y = |-2 - (-4)| = 2$
   $d_{GH}=\sqrt{1^2+2^2}=\sqrt{1+4}=\sqrt{5}\approx2.2$

4. Final answer
   • Distance A-B: $3\sqrt{5} \approx 6.7$
   • Distance C-D: $3\sqrt{5} \approx 6.7$
   • Distance E-F: $\sqrt{17} \approx 4.1$
   • Distance G-H: $\sqrt{5} \approx 2.2$

   Turn 2 Answer

   I'm not really able to generate or draw new images myself, but I can definitely help you figure out where the points go on those grids if you want to sketch them out!

   It looks like you've already got the first one down. Do you want to try plotting Point C (-5, 2) and Point D (-2, -4) next? I can walk through it with you if that helps.

答え

Question Analysis

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Can u turn them to a image

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