數學問題，高中程度，直線

T1.
Slope of MN = (m-1)/(-1-0) = (m-1)/-1 = 1-m.
When m is very very large and postive, slope = tan a = infinity, therefore, a = 90 degree.
When m = 1, slope = 0, therefore tan a = 0, so a = 180 degree.
When m is negative and very large, tan a = infinity, so a = 90 degree.
When m is negative, slope is positive.
Summing up, the range of a is between 0 to 180 degree depending on the value and sign of m.
T2.
Intercept form of straight line is x/a + y/b = 1.
Now a=b , so the line is x + y = a.
Put x=-3 and y=4, a = 1. Therefore, the line is x+y = 1.
T3.
For line ax +by = 1.
x - intercept = 1/a (by putting y = 0)
y-intercept = 1/b (by putting x = 0)
Therefore, area = (1/a)(1/b)(1/2) = 1/(2ab).
T4.
2x + 3y = k..............(1)
x - ky = -12...............(2)
Therefore, the x-coordinate of the interesecting point can be found from the 2 equations by eliminating y.
2kx + 3ky = k^2 ..........(3)
3x - 3ky = -36...............(4)
(3) + (4) we get (3+2k)x = k^2 - 36...........(5)
Since the interesecting point is on the y-axis, x-coordinate is 0.
Put x in (5) to 0, we get k^2 - 36 = 0, therefore, k = +/-6.
T5.
Let slope of line I be tan A = m.
Slope of line 2x + y = 1 is -2 = tan B.
But A + 45 = B
tan(A + 45) = tan B
m/(1-m^2) = -2
m = -2 + 2m^2
2m^2 - m - 2 = 0.
m= (1+sqrt17)/4 and (1-sqrt17)/4 = slope of line I.

2008-07-31 08:31:28 補充：
Correction: Sorry, line 5 and below of T5 is wrong, should be (1+m)/(1-m) = -2. So m = 3.
One more solution is A+ 135 = B, tan(A+135) = B, (m-1)/(1+m) = -2. So m = -1/3.

2008-07-31 08:36:13 補充：
Should be tan(A + 135) = tan B.