Anonymous 發問於 科學及數學數學 · 1 十年前

Intersection point

the equations of two sides of a parallelogram ABCD are

AB: 2x + 3y - 4 = 0

AD: x - y + 3 = 0

The coordinates of C is (4, 2)

Find the coordinates of A, B and D.

1 個解答

評分
  • Marie
    Lv 6
    1 十年前
    最愛解答

    Point A is the intersection point of AB and AD

    AB: 2x + 3y - 4 = 0 .... [1]

    AD: x - y + 3 = 0

    AD: x = y - 3 .... [2]

    substitute [2] into [1]:

    2(y - 3) + 3y - 4 = 0

    2y - 6 + 3y - 4 = 0

    5y = 10

    y = 2

    then x = y - 3 = -1

    A (-1, 2)

    Let b be the x-coordinate of Point B

    put x = b into [1]

    2b + 3y - 4 = 0

    3y = 4 -2b

    y = (4 - 2b)/3

    B (b, (4 - 2b)/3)

    Slope of AD = slope of BC = 1

    1 = [(4 - 2b)/3 - 2] / (b - 4)

    b - 4 = (4 - 2b)/3 - 2

    3(b - 2) = 4 - 2b

    3b - 6 = 4 - 2b

    5b = 10

    b = 2

    (4 - 2b)/3 = (4 - 4) / 3 = 0

    B (2, 0)

    Let d be the x-coordinate of Point D

    put x = d into [2]

    x = y - 3

    y = d + 3

    D (d, d + 3)

    Slope of AB = slope of DC = -2/3

    -2/3 = [d + 3 - 2] / (d - 4)

    -2(d - 4) = 3(d + 1)

    -2d + 8 = 3d + 3

    5d = 5

    d = 1

    d + 3 = 4

    D (1, 4)

還有問題嗎?立即提問即可得到解答。