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anonymous

  • one year ago

The diagram shows a rectangle ABCD. The point A is (2,14). B is (-2,8) and C lies on the x-axis (?,0) Find the equation of BC. the coordinates of C and D.

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  1. anonymous
    • one year ago
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    One second, I have done some work, just trying to find the coordinates for C and D. Missing X values.

  2. anonymous
    • one year ago
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    y-8= -2/3 (x+2) for BC equation. C (?,0) D (?,6) Not sure what to do. I did slope equation. 14-8/2+2 = 6/4 3/2 then opp slope recip. -2/3

  3. anonymous
    • one year ago
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    okay @.@ may i ask what are the question that you want to ask @.@

  4. anonymous
    • one year ago
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    How to find the x values for those coordinates. :p

  5. anonymous
    • one year ago
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    okay can you show me how you got those answers up there? cuz i confuse by your own answers and question @.@

  6. anonymous
    • one year ago
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    Apologies, I have a packet with an image. My answers are accurate, which I am completely confident of.

  7. anonymous
    • one year ago
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    @iYuko

  8. jim_thompson5910
    • one year ago
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    The equation of line BC, in point slope form, is \(\Large y - 8 = -\frac{2}{3}(x+2)\). In slope-intercept form, the equation is \(\Large y= -\frac{2}{3}x+\frac{20}{3}\) It doesn't matter which equation you use. Plug in y = 0 and solve for x to get the x coordinate of point C.

  9. anonymous
    • one year ago
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    Can you explain to me how you got 20/3? @jim_thompson5910

  10. anonymous
    • one year ago
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    Oh, I see, for some reason I attempted to use outside numbers to distribute, I completely understand what I did wrong, thank you so much!

  11. jim_thompson5910
    • one year ago
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    Sure. Here are the steps I did to convert from point slope form to slope intercept form. \[\Large y - 8 = -\frac{2}{3}(x+2)\] \[\Large y - 8 = -\frac{2}{3}(x)-\frac{2}{3}(2)\] \[\Large y - 8 = -\frac{2}{3}x-\frac{4}{3}\] \[\Large y - 8+8 = -\frac{2}{3}x-\frac{4}{3}+8\] \[\Large y = -\frac{2}{3}x-\frac{4}{3}+8\] \[\Large y = -\frac{2}{3}x-\frac{4}{3}+8*\frac{3}{3}\] \[\Large y = -\frac{2}{3}x-\frac{4}{3}+\frac{24}{3}\] \[\Large y = -\frac{2}{3}x+\frac{-4+24}{3}\] \[\Large y = -\frac{2}{3}x+\frac{20}{3}\]

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