a-maths locus

A variable line through the point C(3,4) cut the x-axis and y-axis at A and B respectively. Through A and B , perpendiculars are drawn to the x -axis and the y-axis respectively.The perpendiculars meet at the point M. Find the equation of the locus of M?
過C點的直線為(m為直線的斜率，為一變數)。
y = m(x – 3) + 4
則
A點的x坐標為
0 = m(x + 3) + 4
mx + 3m + 4 = 0 ___(1)
B點的y坐標為
y = m(0 + 3) + 4
y = 3m + 4
由這式
m = (y – 4)/3 將 m代入(1)式
[(y – 4)/3]x + 3[(y – 4)/3] + 4 = 0
全式乘3
(y – 4)x + 3(y – 4) + 12 = 0
yx – 4x + 3y – 12 + 12 = 0
yx – 4x + 3y = 0
so Find the equation of the locus of M is
yx – 4x + 3y = 0

X-ordinate of M is the x-intercept of the line. And similar to Y-ordinate is the y-intercept.

Let m be the slope of the line.
Then the equation of the line is:

y-4 = m(x-3)
when x=0, y = m3/4 (y-intercept)
when y=0, x = m4/3 (x-intercept)

Eliminate m from above equation,

4/3 * y = m = 3/4*x

16y = 9x
0 = 9x -16y is the locus of M.

2006-11-12 00:25:01 補充：
My answer is wrong. Do not use it.