直线l:y=-3/4x-2/3沿x轴翻折,得到一条新直线与x轴交于点A,与y轴交于点B

1个回答

  • (1) y = -3x/4 - 3/2

    y = 0, x = -2, A(-2, 0)

    x = 0, y = -3/2, B'(0, -3/2)

    沿x轴翻折, B'变为B(0, 3/2)

    直线AB的解析式: x/(-2) + y/(3/2) = 1

    y = 3x/4 + 3/2

    (2)

    C1: y = 2x²/3, 对称轴为x = 0

    设向右平移a, C2: y = 2(x - a)²/3

    C2对称轴为x = a

    x = 0, y = 2a²/3

    直线y = 3x/4 + 3/2上, y = 2a²/3, x = 8a²/9 - 2

    F(8a²/9 - 2, 2a²/3)

    C2对称轴为: x = (8a²/9 - 2 + 0)/2 = 4a²/9 - 1 = a

    4a² - 9a - 9 = 0

    (4a + 3)(a - 3) = 0

    a = -3/4或a = 3

    C2: y = 2(x - 3)²/3 = 2x²/3 -4x + 6

    y = 2(x + 3/4)²/3 = 2x²/3 + x + 3/8

    (3)

    F在y轴右侧, a = 3, F(6, 6)

    G(6, 0)

    H(6, -6)

    AH = AF = √[(-2 - 6)² + (0 - 6)²] = 10

    HF = 12

    △AFH的周长L = 10*2 + 12 = 32

    △AFH的面积S = (1/2)HF*AG = (1/2)*(6 + 6)*(6 + 2) = 48

    M(m, 3m/4 + 3/2), N(6, n)

    △MNF的面积S' = S/2 = 24 = (1/2)(M与HF的距离)*NF

    = (1/2)(6 - m)(6 - n)

    (6 - m)(6 - n) = 48 (i)

    AM² = (m+ 2)² + (3m/4 + 3/2 - 0)² = (25/9)(m + 2)²

    AM = 5(m + 2)/4

    HN = n + 6

    AM + HN + AH = L/2

    5(m + 2)/4 + n + 6 + 10 = 32/2 (ii)

    联立(i)(ii): 5m² + 4m - 12 = 0

    m = 6/5或m = -2 (点A, 舍去)

    M(6/5, 12/5)

    n = -4, N(6, -4)

    直线m的解析式: (y + 4)/(12/5 + 4) = (x- 6)/(6/5 - 6)

    y = -4x/3 + 4