Sunday

September 21, 2014

September 21, 2014

Posted by **Anonymous** on Sunday, May 4, 2008 at 2:12pm.

(a)Calculate the angle ABC

(b)Show that 3x+4y-19=0 is an equation for line L which passes through A and B

(c)Find a vector equation for line L

(d)(i)give an equation for the circle with the line segment AB as diameter

(ii)Give an equation for the tangent to the circle at the point B

(e)(i)Find the distance from O,the origin,to line L

(ii)Give an equation for the line through O and perpendicular to L

(iii)Find an equation for the smallest circle through circle O which has a segment of L as a diameter

(f)Calculate the area of the triangle

ABC

I need mostly need help on c,d,e

- Geometry -
**Damon**, Sunday, May 4, 2008 at 4:43pmVector 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

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