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anonymous

  • 5 years ago

I am trying to find an equation using the give pair of points (1/4,-1/2) and (3/4,3)

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  1. anonymous
    • 5 years ago
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    You need to use the two-point formula for a line. It says, if you have two points,\[(x_1,y_1)\]and\[(x_2,y_2)\]then the equation of the line containing those points is\[y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)\]You can make any one of your two points be (x1,y1). The other will have to be the (x2,y2). So here, you'd have\[y-(-\frac{1}{2})=\frac{3-(-1/2)}{3/4-1}(x-1/4)\]that is\[y+1/2=\frac{7/2}{-1/4}(x-1/4) \rightarrow y = -14(x-1/4)-1/2\]

  2. anonymous
    • 5 years ago
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    You can expand to get\[y=-14x+7/2-1/2=-14x+3\]So\[y=-14x+3\]

  3. anonymous
    • 5 years ago
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    we are suppossed to use the mx+b formula is what it says. Is that possible?

  4. anonymous
    • 5 years ago
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    Sure. You should have said so! :) I'll do it differently.

  5. anonymous
    • 5 years ago
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    Well, actually, if you notice when I gave you the two-point formula, outside the \[(x-x_1)\]part there is\[\frac{y_2-y_1}{x_2-x_1}\]...that's the slope, m. So you can calculate that first, then you pick any one of the two points to do the rest\[y-y_1=m(x-x_1)\]

  6. anonymous
    • 5 years ago
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    They're actually the same formula...you get m from the two points.

  7. anonymous
    • 5 years ago
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    You're not actually told the intercept in your question, so can't read off 'b'. You would need to have one of your points, (0,b).

  8. anonymous
    • 5 years ago
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    So, if you were given two points, (1,2) and (0,3), you could use y=mx+b as: m= (3-2)/(0-1)=-1 b=3 so y=-x+3

  9. anonymous
    • 5 years ago
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    Your question's slightly different; you can't jump to y=mx+b...you need to find b from the points like I've done.

  10. anonymous
    • 5 years ago
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    the only points i have are (1/4,-1/2) and (3/4,3) sorry i did not mention it before. At the bottom it says what is the equation of the line? y= (Simplify your answer.Type the answer in the form y=mx+b using intergers or fractions) this is why I am having so much problem trying to figure out what they want me to do because everytime I put in an answer they say it is wrong

  11. anonymous
    • 5 years ago
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    Okay...they're saying "Find the equation of the line ANY way you like (from ways you're taught) and THEN put it in the form y=mx+b...then type that in." I'll do it on paper to double-check.

  12. anonymous
    • 5 years ago
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    okay- I made a mistake before - when I type on the fly, I sometimes stuff up. The equation of your line in the form you need is\[y=7x-\frac{9}{4}\]

  13. anonymous
    • 5 years ago
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    ok that was right that time. thanks

  14. anonymous
    • 5 years ago
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    np

  15. anonymous
    • 5 years ago
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    ok now I have to write an equation of the line containing the given point and parallel to the given line. express your answer in the form of y=mx+b (6,7);x+7y=2

  16. anonymous
    • 5 years ago
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    Well, the second line needs to be parallel to the first, so it must have the same slope. You should arrange the first equation in the form y=mx+b and then 'read off' the slope. That will be the slope of the second line. Now, since you have a point that the second line must have, and the slope, you can use the point-slope formula I gave above y - y1 = m(x - x1) You expand and rewrite it as a last step in the form you need.

  17. anonymous
    • 5 years ago
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    ok so the slope would be 1 then right?

  18. anonymous
    • 5 years ago
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    No...you have to put the equation in the right form. x+7y=2 is NOT in the form y=mx+_b

  19. anonymous
    • 5 years ago
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    \[y=-\frac{1}{7}x+\frac{2}{7}\]

  20. anonymous
    • 5 years ago
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    So the slope is -1/7

  21. anonymous
    • 5 years ago
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    Your new line will be\[y-y_1=m(x-x_1) \rightarrow y-7=-\frac{1}{7}(x-6)\]

  22. anonymous
    • 5 years ago
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    \[y=-\frac{1}{7}x+\frac{55}{7}\]when you rearrange and put it into the form you need.

  23. anonymous
    • 5 years ago
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    so would I do the same thig with these points (-2.5);5x=7y+8

  24. anonymous
    • 5 years ago
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    What are you being asked to do?

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