Amath questions

(a)
PQ: 3x + y - 2 = 0 ...... (1)
PS: 5x - 3y - 8 = 0 ...... (2)

PQ meets PS at P.

(1)x3:
9x + 3y - 6 = 0 ...... (3)

(2)+(3):
14x -14 = 0
14x = 14
x = 1

Put x = 1 into (1):
3(1) + y - 2 = 0
y + 1 = 0
y = -1

Hence, the coordinates of P is (1, -1).



Let (a, b) be the coordinates of R.

The diagonals of a parallelogram bisect each other.
(4, -7) is the mid-point of the diagonal PR.

(1 + a)/2 = 4
1 + a = 8
a = 7

(-1 + b)/2 = -7
-1 + b = -14
b = -13

Hence, the coordinates of R is (7, -13).

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(b)
PS: 5x - 3y - 8 = 0
Slope of PS = -5/(-3) = 5/3
QR//PS
Slope of QR = Slope of PS = 5/3

QR passes through R(7, -13) with slope 5/3.
The point-slope form of QR:
y + 13 = (5/3)(x - 7)
3(y + 13) = 5(x - 7)
3y + 39 = 5x - 35
QR: 5x - 3y - 74 = 0


PQ: 3x + y - 2 = 0
Slope of PQ = -3/1 = -3
RS//PQ
Slope of RS = Slope of PQ = -3

RS passes through R(7, -13) with slope -3.
The point-slope form of QR:
y + 13 = (-3)(x - 7)
y + 13 = -3x + 21
QR: 3x + y - 8 = 0
=

2008-11-23 17:37:14 補充：
Typing mistake: The last line should be

PS: 3x + y - 8 = 0