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anonymous

  • one year ago

A line passes through point (3, 7) and has a slope of 3/4

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  1. anonymous
    • one year ago
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    @DaBest21 @harryhart

  2. pooja195
    • one year ago
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    in what form?

  3. anonymous
    • one year ago
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    y=mx+b slope y-ntercept

  4. anonymous
    • one year ago
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    Let the required line be y = mx+c in the slope intercept form, where m is the slope and c is the y intercept. We determine m and c by the using given conditions. m = 3/4, given. So y = (3/4)x+c (1). The line at (1) has the point (-8, 4) on it. Therefore 4 = (3/4)*(-8)+c...(2) (1)-(2) gives: y-4 = (3/4)(x+8). => y -4 = (3/4)x + 6 We multiply 4 to get integral coefficients. 4(y-4) = 3x+24 We rearrange. 3x-4y+24+16= 0 3x-4y+40 = 0 is the required line.

  5. anonymous
    • one year ago
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    the options are- y=3/4x+19/4 and y=3x-19

  6. anonymous
    • one year ago
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    what do you think it would be the answer?

  7. anonymous
    • one year ago
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    y=3/4x+19/4

  8. anonymous
    • one year ago
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    but im not sure

  9. pooja195
    • one year ago
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    An easier approach for me atlest is this way: \[\huge y-y_1=m(x-x_1)\] where m=slope (x1,y1) (3, 7) and has a slope of 3/4 y-7=3/4(x-3) from here after you distribute and solve for y you get y=3/4x+19/4

  10. pooja195
    • one year ago
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    So what you said is correct

  11. anonymous
    • one year ago
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    sweet

  12. anonymous
    • one year ago
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    If point A(x, 5) lies on the line, the value of x is 33 or 1/3

  13. anonymous
    • one year ago
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    still need help @harryhart

  14. anonymous
    • one year ago
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    @DaBest21

  15. anonymous
    • one year ago
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    open a new question

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