show that the triangle with vertices at A(0,2), B(3,0) and C(4,8) is a right triangle?

using phytagorean theorem

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  • 8 years ago
    Favorite Answer

    Since phytagorean theorem to be used , calculate basic AB = √ ((0-3) √ ((0-3)² + (2-0)² ) =√13

    BC = √ ((3-4)²+ (0-8)² ) =√65 , AC = √ ((0-4)² + (2-8)² ) =√52

    Using phytagorean theorem , AB² + AC² =(√13)² + (√52)² = 65 = BC²

    It is a Right angle triangle with , Hypotunuse BC and A = 90°

  • 8 years ago

    A (0 ; 2) B (3 ; 0) C (4 ; 8)

    Distance AB

    xAB = xB - xA = 3 - 0 = 3

    yAB = yB - yA = 0 - 2 = - 2

    AB² = xAB² + yAB² = (3)² + (- 2)² = 9 + 4 = 13

    Distance BC

    xBC = xC - xB = 4 - 3 = 1

    yBC = yC - yB = 8 - 0 = 8

    BC² = xBC² + yBC² = (1)² + (8)² = 9 + 64 = 65

    Distance AC

    xAC = xC - xA = 4 - 0 = 4

    yAC = yC - yA = 8 - 2 = 6

    AC² = xAC² + yAC² = (4)² + (6)² = 16 + 36 = 52

    You can see that:

    13 + 52 = 65

    AB² + AC² = BC² → the triangle is right @ the point A

  • Glipp
    Lv 7
    8 years ago

    vector AB = (3, -2)

    vector BC = (1, 8)

    vector AC = (4, 6)

    AB . AC = 3*4 + (-2)*6 = 0 (dot product)

    so there is no projection of AB onto AC so AB is at right angles to AC

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