Posted by Anonymous on Sunday, May 4, 2008 at 2:12pm.
Vector BA = -4i + 3j
Vector BC = -1i + 7j
|BA| = sqrt (16+9) = 5
|BC| = sqrt (1 + 49) = sqrt (50)=5sqrt2
BA dot BC = 4+21 = 25
BA dot BC = |BA| |BC| cos ABC
so
25 = 5 * 5 sqrt 2 * cos ABC
cos ABC = sqrt 2 / 3 = .707
ABC = 45 degrees
b
L is y = -(3/4) x + 19/4
note slope of vector AB = -3/4
now does L go through A which is (1,4)?
4 = -(3/4)x + 19/4
16 = -3x + 19
x = 1 sure enough L goes through A and B
c
did that right off in A but I did BA = -4i + 3 j
we want AB = 4i -3j = -BA
d
center of circle at center of AB
x = (5+1)/2 = 3
y = (4+1)/2 = 2.5
radius = |BA|/2 = 2.5
so
(x-3)^2 +(y-2.5)^2 = 2.5^2 = 6.25
tangent at B has slope perpendicular to L -1/-(3/4) or 4/3 and goes through B
y = (4/3) x + b
1 = (4/3)5 + b
b = 1 - 20/3 = -17/3
so y = (4/3) x -17/3
or 3 y = 4 x - 17
e
call line perpendicular to L through the origin P
equation of P is
y = m' x + 0 (because through origin b =0)
m' = -1/slope of L = 4/3
so
y = (4/3)x is equation of P (that is e ii by the way
where does L hit P ?
(4/3) x = -(3/4) x + 19/4
(4/3 + 3/4) x = 19/4
(16 + 9 ) x/12 = 19/4
25 x = 3*19
x = 3*19/25
y = 4*19/25
the distance from O to that intersection is (19/25)sqrt (3^2+4^2) =(19/25)*5 = 19/5 that is e i answer
e iii
well I know center is at x = 3*19/25 and y = 4*19/25 and the radius is 19/5
so
(x - 3*19/25)^2 + (y - 4*19/25)^2 = 19^2/25^2
f
area = (1/2) |BA||BC| sin ABC
= (1/2)(5)(5 sqrt 2)(sin45)
= (25/2) sqrt 2 * 1/sqrt 2
= 25/2 = 12.5