Maths(Coordinate Treatment of Simple Locus Problems)

(a) L: x+3y-2+k(x-y-6) = 0
x+3y-2+k(x-y-6) = 0
x + 3y - 2 + kx - ky - 6k = 0
ky - 3y = kx + x - 6k - 2
(k - 3)y = (k + 1)x + - 2(3k +1) =0
y = [(k + 1)/(k - 3)]x - [2(3k +1)/(k - 3)]
therefore, slope of L = (k + 1)/(k - 3)
(b) the line x+2y-4=0 can be expressed as:
2y = -x + 4
y = -(1/2)x + (4/2)
therefore slope of the line x+2y-4=0 is -1/2
if L is parallel to the line x+2y-4=0
slope of L = -1/2
(k + 1)/(k - 3) = -1/2
2k + 2 = -k + 3
3k = 1
k = 1/3

(c) L can be written as (k+ 1)x + (3 -k)y - 2(3k + 1) = 0
Let the coordinates of P be (a, b), and substitue (a,b) into L, it becomes
(k+ 1)a + (3 - k)b - 2(3k + 1) = 0 .................(1)

L passes through a fixed point P for any value of k, so
put k = -1, (1) becomes
(-1 + 1)a + [3-(-1)]b - 2[3(-1) + 1] = 0
4b + 4 = 0
b = -1

put k = 3, (1) becomes
(3+ 1)a + (3 - 3)b - 2[3(3) + 1] = 0
4a - 20 = 0
a = 5

therefore, the coordinates of P is (5, -1)

The equation of line L is x+3y-2+k(x-y-6)=0
(a)Express the slope of L interms of k.
由L:x+3y-2+k(x-y-6)=0
x+3y-2+kx-ky-6k=0
ky-3y=kx+x-(6k+2)
(k-3)y=(k+1)x-(6k+2)
y=[(k+1)/(k-3)]x-(6k+2)/(k-3)
即L的斜率為(k+1)/(k-3)

(b)FInd the value of k if L is parallel to x+2y-4=0.
∵L is parallel to x+2y-4=0
x+2y-4=0的斜率為-1/2
即(k+1)/(k-3)=-1/2
2k+2=-k+3
3k=1
∴k=1/3

(c)If L passes through a fixed point P for any value of k,find the coorsinates of P
由L : x+3y-2+k(x-y-6)=0與k無關得
{x+3y-2=0---------(1)
{x-y-6=0------------(2)
(1)-(2)
4y+4=0
y=-1
代入(2)
x+1-6=0
x=5
∴P的坐標為(5,-1)