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nasty-shy [4]
3 years ago
14

Line JK passes through points J(–3, 11) and K(1, –3). What is the equation of line JK in standard form?

Mathematics
1 answer:
nikitadnepr [17]3 years ago
6 0
You can start with the form
  ∆y(x -x1) -∆x(y -y1) = 0
Here, we have
  ∆y = 11-(-3) = 14
  ∆x = -3-1 = -4
and we can choose (x1, y1) = (1, -3). This gives
  14(x -1) -(-4)(y -(-3)) = 0
  14x +4y -2 = 0
All these terms have a common factor of 2 that we can remove. Adding 1 to the result puts it in standard form:

  7x +2y = 1

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Step-by-step explanation:

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Step-by-step explanation: I had a similar project so I used mine as an example and I rewrote it. Hope this helps

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5 0
3 years ago
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V125BC [204]

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Step-by-step explanation:

6 0
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Read 2 more answers
Giving brainlist easy math problem!
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Answer:

b = 10

-2d = -38

x = -1.2

Step-by-step explanation:

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7 0
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