Locus help (1 Viewer)

[Wooz]

Could someone fully explain the working in this question, thanks in advance.

I really need help with this question i've drawn a diagram and found the gradient of one of the bisects and one of the midpoints finding the eqn of one of the perpendicular lines being y=-x+4.

Question: A(-6,4), B(4,4) and C(0,0) are the verticies of a triangle. Find the eqn's of the perpendicular bisectors of each of the sides of this triangle. Find the pt. where the perpendiculoar bisectors of AB and AC meet and shwo that the perrpendicular bisectors of BC also passes thourgh the this point.
If the poing where the perpendicular bisectors meet is calles R, find the length of RA and find the eqn of the circle with centre R and radius RA.
Show by substitution that pionts B and C lie on this circle.

Note: this circle is called the circumcircle of the triangle ABC.

Answer: (-1,5), (x+1)^2 + (y-5)^2 = 26.

 

[ssglain]

A(-6,4), B(4,4) and C(0,0)


First, let the midpoints of AB, AC, BC respectively be L, M, N.
The coordinates of these points are easy to obtain using the midpoint formula.
L(-1, 0)
M(-3, 2)
N(2, 2)

Next, find the gradients of sides using the gradient formula.
Gradient of AB = 0
Gradient of AC = -2/3
Gradient of BC = 1

Using these find the gradient of their respective perpendicular bisectors. (Remember that if two lines are perpendicular, then the product of their gradients is -1.) Then use the point-gradient formula to find their equations.
The p.b. of AB is a vertical line through L(-1, 0) because AB is horizontal (grad=0).
Therefore, p.b. AB: x = -1
The p.b. of AC has gradient 3/2 and passess through M(-3, 2).
Therefore, p.b. AC: y - 2 = (3/2)*(x + 3), i.e. y = (3/2)x + (13/2)
The p.b. of BC has gradient -1 and passes through N(2, 2).
Therefore, p.b. BC: y - 2 = -1*(x - 2), i.e. y = -x + 4

Now solve for the point of intersection, R, of p.b. AB and p.b. AC by substituting x = -1 into y = (3/2)x + (13/2).
y = (3/2)*(-1) + (13/2) = 5
Therefore, R(-1, 5).

Verify that R lies on p.b. BC by sustitution:
LHS = 5
RHS = -(-1) + 4 = 5 = LHS
Therefore BC also passes through R(-1, 5)

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Now let's deal with the circle.

Find the length of AR using the distance formula. AR = SQRT(26).
Remember that the standard equation of a circle with centre (h, k) and radius r is (x - h)^2 + (y - k)^2 = r^2.
Therefore, a circle with centre R(-1, 5) and radius AR = SQRT(26) has the equation: (x + 1)^2 + (y - 5)^2 = 26

Verify that B(4, 4) and C(0, 0) lie on the circle by substitution:
For B(4, 4): LHS = 5^2 + (-1)^2 = 26 = RHS
For C(0,0): LHS = 1^2 + (-5)^2 = 26 = RHS

 

[Wooz]

Cheers,