# Maths(Coordinate Treatment of Simple Locus Problems)

The equation of line L is x+3y-2+k(x-y-6)=0

(a)Express the slope of L interms of k.

(b)FInd the value of k if L is parallel to x+2y-4=0.

(c)If L passes through a fixed point P for any value of k,find the coorsinates of P.

### 2 個解答

• 最佳解答

(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)