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robinfr93 Group Title

Let PQR be a right angled isosceles triangle in a cartesian plane, right angled at P(2,1). If the equation of the line QR is 2x+y = 3, then the equation representing the pair PQ and PR is..

  • 2 years ago
  • 2 years ago

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  1. ChmE Group Title
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    Draw a picture first. it always helps visualize

    • 2 years ago
  2. robinfr93 Group Title
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    Well there wasn't one but neway heres what I can do..

    • 2 years ago
  3. ChmE Group Title
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    When it says right angled at (2,1) I think at that point it is 90 deg

    • 2 years ago
  4. robinfr93 Group Title
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    |dw:1347994216890:dw|

    • 2 years ago
  5. robinfr93 Group Title
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    yes dats pretty obvious isn't it??

    • 2 years ago
  6. ChmE Group Title
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    Right angle isoceles means that the sides to each side of the right angle are equal. Ya I'd say so after the picture.

    • 2 years ago
  7. robinfr93 Group Title
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    yup, both the angles will be 45 degrees, i.e, 45+45+90

    • 2 years ago
  8. robinfr93 Group Title
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    well ChmE??

    • 2 years ago
  9. robinfr93 Group Title
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    well @ChmE

    • 2 years ago
  10. robinfr93 Group Title
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    Can't anybody solve this??

    • 2 years ago
  11. bahrom7893 Group Title
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    Ok let me try.

    • 2 years ago
  12. bahrom7893 Group Title
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    I want to say PQ is just x=2 and PR is just y=1, but what if the triangle is at an angle..

    • 2 years ago
  13. TuringTest Group Title
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    just find points Q and R, given that we know P

    • 2 years ago
  14. TuringTest Group Title
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    then use point slope and all that business

    • 2 years ago
  15. robinfr93 Group Title
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    Its an isosceles right angled triangle that means 2 of its angle are 45 degrees and one is at 90 degrees..

    • 2 years ago
  16. robinfr93 Group Title
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    @TuringTest the question is HOW???

    • 2 years ago
  17. TuringTest Group Title
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    let's draw a bigger pic

    • 2 years ago
  18. bahrom7893 Group Title
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    |dw:1347994331965:dw|

    • 2 years ago
  19. robinfr93 Group Title
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    btw the answers ae PQ = x+3y -5 and PR = 3x-y-5

    • 2 years ago
  20. bahrom7893 Group Title
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    lol does that help?

    • 2 years ago
  21. robinfr93 Group Title
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    @bahrom7893 No dude!!! Its already in the question... :P

    • 2 years ago
  22. TuringTest Group Title
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    |dw:1347994441088:dw|

    • 2 years ago
  23. robinfr93 Group Title
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    Yeah now what??

    • 2 years ago
  24. robinfr93 Group Title
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    I got it so far!!! next step will be what??

    • 2 years ago
  25. TuringTest Group Title
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    at x=2 what is y ?

    • 2 years ago
  26. TuringTest Group Title
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    calm down dude I am explaining...

    • 2 years ago
  27. robinfr93 Group Title
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    :P Sorry its just that its getting in my head bad!!

    • 2 years ago
  28. robinfr93 Group Title
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    well all that is given in the ques, no additionally details are provided..

    • 2 years ago
  29. TuringTest Group Title
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    though I must admit I;m not sure exactly what they are asking the equation for the pair? what do they mean?

    • 2 years ago
  30. robinfr93 Group Title
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    hahhahahahah.. Exactly what Ive stumbled upon..

    • 2 years ago
  31. robinfr93 Group Title
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    by pair of lines they mean PQ and PR

    • 2 years ago
  32. TuringTest Group Title
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    but infinitely I assume, not a finite line segment....

    • 2 years ago
  33. robinfr93 Group Title
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    nope its finite alright!!!

    • 2 years ago
  34. TuringTest Group Title
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    then the answer must give bounds to x and y

    • 2 years ago
  35. robinfr93 Group Title
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    Look will it help if I write the solution down, The thing is Ive got the solution but I don't really understand it very well...

    • 2 years ago
  36. TuringTest Group Title
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    |dw:1347994766490:dw|

    • 2 years ago
  37. TuringTest Group Title
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    yeah that may help me understand what they are asking

    • 2 years ago
  38. robinfr93 Group Title
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    The equations of PQ and Pr are... y-2 = [-2 -/+ tan 45/1+/- (-2)tan 45] * (x-2) <=> HOW??? Solving this they Got PQ ad PR...

    • 2 years ago
  39. robinfr93 Group Title
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    PR*

    • 2 years ago
  40. TuringTest Group Title
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    oh I misinterpreted the problem I see

    • 2 years ago
  41. TuringTest Group Title
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    |dw:1347994998473:dw|so my assumption that the legs of the triangle followed the grid was wrong

    • 2 years ago
  42. robinfr93 Group Title
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    @TuringTest Still don't get...

    • 2 years ago
  43. TuringTest Group Title
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    well now the problem is harder than I thought, so let me think on it I was eating lunch as I was doing this because I thought it would be easy lol

    • 2 years ago
  44. robinfr93 Group Title
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    :P well it isn't @hartnn DUDE!!! You were saying u got it ryt?? Will you do the honor of telling me how??

    • 2 years ago
  45. hartnn Group Title
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    even i thought that lines are vertical and horizontal.....but they r not.....even i m stuck...

    • 2 years ago
  46. hartnn Group Title
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    that seems the formula for tan(A+B) ....in the solution u mentioned....do u have some standard formulas in this topic ?

    • 2 years ago
  47. robinfr93 Group Title
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    I know!!! just one , but I don't understand hw to implement it in this one..

    • 2 years ago
  48. robinfr93 Group Title
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    Two lines will be perpendicular to each other if...coeff of x^2 + coeff of y^2 = 0

    • 2 years ago
  49. robinfr93 Group Title
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    but this is a first degreeequation so can't really put that in...

    • 2 years ago
  50. hartnn Group Title
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    yes, that won't help..

    • 2 years ago
  51. robinfr93 Group Title
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    Think people THINK!!!! The person who can solve this will be titled a born genius.. :P

    • 2 years ago
  52. TuringTest Group Title
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    |dw:1347995522447:dw|

    • 2 years ago
  53. TuringTest Group Title
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    I dunno I'll have to drop this for now...

    • 2 years ago
  54. robinfr93 Group Title
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    NOOOOOOOOOOOOOOOOOOOOOOOOOOO!!!! :'(

    • 2 years ago
  55. hartnn Group Title
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    |dw:1347995617810:dw| i don't know what i m doing......

    • 2 years ago
  56. hartnn Group Title
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    6 equations 8 unknowns.

    • 2 years ago
  57. robinfr93 Group Title
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    :P

    • 2 years ago
  58. hartnn Group Title
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    mM=-1--->7th equation.

    • 2 years ago
  59. hartnn Group Title
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    1 more equation and we are done.

    • 2 years ago
  60. robinfr93 Group Title
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    you mean one more equation if was provided??

    • 2 years ago
  61. hartnn Group Title
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    1 more equation WE need to find.

    • 2 years ago
  62. robinfr93 Group Title
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    Take all the time you need, Ill be here for a while....

    • 2 years ago
  63. hartnn Group Title
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    lol, very big equation it is!! PQ=PR and distance formula :P

    • 2 years ago
  64. robinfr93 Group Title
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    :P

    • 2 years ago
  65. hartnn Group Title
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    we getting 8 equations and 8 unknowns but its almost impossible to solve them simultaneously.....there must be another way...

    • 2 years ago
  66. robinfr93 Group Title
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    I typed in the equation and the solution that was given in the text.. The only questions bugging my mind is Y and HOW??

    • 2 years ago
  67. hartnn Group Title
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    and even how y-2 and x-2 ?? y-2 and x-1 maybe.....

    • 2 years ago
  68. hartnn Group Title
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    we can use help of @ganeshie8 here....

    • 2 years ago
  69. robinfr93 Group Title
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    Is he any good??

    • 2 years ago
  70. hartnn Group Title
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    you will see.

    • 2 years ago
  71. hartnn Group Title
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    do u have formulas like this somewhere ? \(\huge y-y_1=\frac{m_1\pm m_2}{1\mp m_1m_2}(x-x_1)\) with this u directly get to answer..............

    • 2 years ago
  72. hartnn Group Title
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    m1 is slope of given line = -2 m2 is slope between lines = tan 45 .............

    • 2 years ago
  73. ganeshie8 Group Title
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    |dw:1347997360648:dw|

    • 2 years ago
  74. ganeshie8 Group Title
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    im trying to work this out by finding out inclination angles for the two lines PR and PQ

    • 2 years ago
  75. ganeshie8 Group Title
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    since we have a point P(1, 2 ) already, we just need to find the inclination angle

    • 2 years ago
  76. ganeshie8 Group Title
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    |dw:1347997747693:dw|

    • 2 years ago
  77. hartnn Group Title
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    nice hint @ganeshie8 , i have some idea now....

    • 2 years ago
  78. hartnn Group Title
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    inclination angles easy to find.........

    • 2 years ago
  79. ganeshie8 Group Title
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    no this is roundabout way... im sure using ur formula we get to the answer much faste.r...

    • 2 years ago
  80. robinfr93 Group Title
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    can someone just elucidate what or how I can solve the problem??

    • 2 years ago
  81. hartnn Group Title
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    |dw:1347997933295:dw|

    • 2 years ago
  82. hartnn Group Title
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    did u get that figure ? @robinfr93

    • 2 years ago
  83. hartnn Group Title
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    |dw:1347998406604:dw| solved!

    • 2 years ago
  84. hartnn Group Title
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    main thing was to find slope of one of the line = 3 using inclination angles....

    • 2 years ago
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