Coordinate Geometry help (1 Viewer)

[jellybelly59]

It is true in general that the perpendicular bisectors of the sides of a triangle are concurrent and that the point of intersection (called the circumcentre) is the centre of a circle through all three vertices (called the circumcircle). prove this result in general by placing vertices at A(2a,0), B(-2a,0) and C(2b,2c) and proceeding as follows:

(a) Find the gradients of AB, BC and CA, and hence find the eqns of the three perpendicular bisectors.

 

[jellybelly59]

bump

 

[tommykins]

It is true in general that the perpendicular bisectors of the sides of a triangle are concurrent and that the point of intersection (called the circumcentre) is the centre of a circle through all three vertices (called the circumcircle). prove this result in general by placing vertices at A(2a,0), B(-2a,0) and C(2b,2c) and proceeding as follows:

(a) Find the gradients of AB, BC and CA, and hence find the eqns of the three perpendicular bisectors.

At first thought it seems like a very tedious question...

dy/dx of AB = rise/run = 0/4a
BC and CA is the same thing.

eqns of the bisectors -
negative recipricol = normal (as the bisectors are perpindicular)
find midpoint of AB, BA and CA (bisectors)
use point gradient formula.

 

[Timothy.Siu]

this is just tedious...

 

[PC]

It is true in general that the perpendicular bisectors of the sides of a triangle are concurrent and that the point of intersection (called the circumcentre) is the centre of a circle through all three vertices (called the circumcircle). prove this result in general by placing vertices at A(2a,0), B(-2a,0) and C(2b,2c) and proceeding as follows:

(a) Find the gradients of AB, BC and CA, and hence find the eqns of the three perpendicular bisectors.

Gradient of AB = (0 – 0)/(2a + 2a) = 0/4a = 0
Gradient of BC = (0 – 2c)/(–2a – 2b) = –2c/–2(a + b) = c/(a + b)
Gradient of AC = (0 – 2c)/(2a – 2b) = –2c/2(a – b) = –c/(a – b)

Midpoint of AB = ( (2a–2a)/2 , (0+0)/2 ) = (0,0)
Midpoint of BC = ( (–2a+2b)/2 , (0+2c)/2 ) = (b–a,c)
Midpoint of AC = ( (2a + 2b)/2 , (0+2c)/2 ) = (a+b,c)

Gradient of perpendicular bisector of AB is undefined.
Gradient of perpendicular bisector of BC is –(a+b)/c.
Gradient of perpendicular bisector of AC is (a–b)/c.

Equation of perpendicular bisector of AB:
x = 0
Equation of perpendicular bisector of BC:
y – c = –(a+b)/c x (x – (b–a) )
cy – c² = –(a + b)(x – b + a)
cy – c² = –(a + b)x + (a + b)b – (a + b)a
cy – c² = –ax – bx + ab + b² – a² – ab
cy – c² = –ax – bx + b² – a²
(a + b)x + cy + a² – b² – c² = 0
Equation of perpendicular bisector of AC:
y – c = (a–b)/c x (x – (a+b) )
cy – c² = (a–b)(x–a–b)
cy – c² = ax – a² – ab – bx + ab + b²
cy – c² = ax – a² – bx + b²
ax – bx – cy – a² + b² + c² = 0
(a – b)x – cy – a² + b² + c² = 0

I wonder what part (b) asks?!?