F5 math problems ( Locus )

1. slope = tan 45 = 1
If the equation is y = mx + c,
therefore y = x + c as m = 1

Put (5, 7) into the equation,
7 = 5 + c
c = 2
so, the equation is y = x + 2

2. slope of AB = (5 - 3) / (2 - 6) = - 0.5

L is perpendicular to AB,
therefore slope of L * slope of AB = -1
slope of L = -1 / slope of AB
= 2

therefore the equation of L is y = 2x + c
Put A(2, 5) into L,
5 = 2(2) + c
5 - 4 = c
c = 1

the equation of c is y = 2x + 1

1. For an inclination 45°, the slope of the line is tan 45° = 1. We use point-slope form.
(y - 7) / (x - 5) = 1
y - 7 = x - 5
Equation: x - y = -2

2. Slope of L: (5 - 3) / (2 - 6) = -2 / -4 = 1/2
Slope of line perpendicular to L = -1 / (1/2) = -2
We apply point-slope form here.
(y - 5) / (x - 2) = -2
y - 5 = -2x + 4
Equation: 2x + y = 9

2007-10-27 01:09:28 補充：
Some amendments to answer 2:Slope of L: (5 - 3) / (2 - 6) = 2/ -4 = -1/2Slope of line perpendicular to L: -1 / (-1/2) = 2Apply point-slope form.(y - 5) / (x - 2) = 2y - 5 = 2x - 4Equation: 2x - y = -1