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

• 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)