Coordinate geometry?

Question

Consider the triangle that is formed by:
A (-2, 5)
B (5, 1)
C (4, 11)

The point where the perpendicular bisectors intersect is known as the circumcenter, because it is the center of the circle that circumscribes the triangle.

Derive the equation of the circle that circumscribes this triangle.

B is actually (6, 1). Sorry about that.

Answer 1

Consider the triangle that is formed by:
A (-2, 5)
B (6, 1)
C (4, 11)

Derive the equation of the circle that circumscribes this triangle.

The center of the circle is the intersection of the perpendicular bisectors of the sides of the triangle.

The midpoint M1 of A (-2, 5) and B (6, 1) is:

M1[(-2+6)/2, (5+1)/2] = M1(2, 3)

The slope m1 of AB = ∆y/∆x = (1-5)/(6+2) = -4/8 = -1/2. The slope of the perpendicular bisector m' is the negative reciprocal, m1' = -1/m1 = 2.

The equation of the perpendicular bisector of AB is

y - 3 = 2(x - 2) = 2x - 4
y = 2x - 1
________________

The midpoint M2 of A (-2, 5) and C (4, 11) is:

M2[(-2+4)/2, (5+11)/2] = M2(1, 8)

The slope m2 of AC = ∆y/∆x = (11-5)/(4+2) = 6/6 = 1. The slope of the perpendicular bisector m2' is the negative reciprocal, m2' = -1/m2 = -1.

The equation of the perpendicular bisector of AC is

y - 8 = -(x - 1) = - x + 1
y = -x + 9
__________________

Find the intersection of the two perpendicular bisectors.

y = 2x - 1
y = -x + 9

2x - 1 = -x + 9
3x = 10
x = 10/3

y = 2x - 1 = 2(10/3) - 1 = 20/3 - 1 = 17/3

The center of the circle (h,k) = (10/3, 17/3).

The radius is the distance from the center of the circle to any one of the points. Let's choose B(6, 1).

radius = r = √[(10/3 - 6)² + (17/3 - 1)²] = √(64/9 - 196/9)

r = √(260/9)

The equation of the circle is that circumscribes the triangle is:

(x - h)² + (y - k)² = r²

(x - 10/3)² + (y - 17/3)² = 260/9

Answer 2

o.ok. i will provide you with "some" help, yet you may want to do the artwork your self. a million. you've a quadrilateral with particular vertices. you want to instruct it really is an isosceles trapazoid. What did you study an isosceles trapazoid??? nicely, you comprehend that: a million. the bases are parallel 2. the perimeters are equivalent 3. the "base" is longer than the "proper" What do you ought to artwork with?? only the co-ordinate factors. nicely, it 'pears to ME that you could use the area formula to discover the dimensions of the perimeters and instruct that they are equivalent. you could then discover the slope of the bottom and proper to instruct they are parallel. then you fairly can discover the dimensions of the bottom and proper to instruct they are UNEQUAL and for that reason you do not have a parallelogram. issue #2. What are the features of a sq.. nicely, if I remember properly, it really is 4 aspects equivalent and one perfect attitude. for that reason, you should use the area formula to instruct all the perimeters are equivalent. THEN, you should use the area formula to instruct one diagonal is the sum of the squares of two aspects, for that reason proving the protected attitude is a perfect attitude. do not make those products harder than it really is. continuously, tony