附加數,線性方程

(a) Mid - point of AB is (-1/2, 3/2).
Slope of AB = (2-1)/(3 + 4) = 1/7.
So equation of perpendicular bisector is
y - 3/2 = 1/7( x + 1/2)
14y - 21 = 2x + 1
14y = 2x + 22.
(b) Let the x - coordinate of the point P be x
so y-coordinate is 2x + 3.
Since AP = BP
so (x - 3)^2 + (2x + 3 - 2)^2 = ( x + 4)^2 + ( 2x + 3 -1)^2
x^2 + 9 - 6x + 4x^2 + 1 + 4x = x^2 + 16 + 8x + 4x^2 + 4 + 8x
10 - 2x = 16x + 20
-10 = 18x, x = -5/9, so y = 2(-5/9) + 3 = 17/9.
so the point P is ( -5/9, 17/9).
(a)
Slope of line x + y - 1 = 0 is -1 = tan a
Slope of line x - 2y - 2 = 0 is 1/2 = tan b
Let the obtuse angle between the lines be k
so b + k = a
k = a - b
tan k = tan ( a - b) = (tan a - tan b)/(1 + tan a tan b)
= (- 1 - 1/2 )/( 1 - 1/2) = (-3/2)/(1/2) = - 3.
so k = arctan (-3) = 108.43 degree
so the acute angle between the lines is 180 - 108.43 = 71.57 degree.
(b)
Let the angle L3 makes with the x - axis be m
so b + 71.57 = m
tan m = tan ( b + 71.57) = (tan b + tan 71.57)/( 1 - tan b tan 71.57)
= (1/2 + 3)/( 1 - 3/2) = (7/2)/(-1/2) = -7 = slope of L3.
Since L3 passes through (-2, 0), so equation of l3 is
y = -7(x + 2) = - 7x - 14
or 7x + y + 14 = 0.