A(-3,-4) , B(-5,0) , C(3,0) are the Vertices of Triangle ABC.Find The co-ordinate of the circumcentre of Triangle ABC
Answers
Answer:
The co-ordinates of the circumcenter of triangle ABC is P (1, \frac{1}{2}1,
2
1
)
Step-by-step explanation:
Given Data
The vertices of given triangle are A (-3,-4), B (-5,0) and C (3,0)
To find - the circumcentre P(x,y) of the triangle ABC.
The point of intersection of all the three perpendicular bisectors of the sides of triangle is called as Circumcenter of a triangle.
PA² = PB² = PC²
Lets take PA² = PB²
Where P = (x,y) , A = (-3,-4) and B = (-5,0)
(x + 3)² + (y +4)² = (x +5)² + (y - 0)²
x² + 9 + 6x + y² + 16 + 8y = x² + 25 + 10x + y²
Cancel x² and y² on both sides of the above equation
9 + 6x + 16 + 8y = 25 + 10x
25 + 8y = 25 + 4x
8y = 4x
y = \frac{1}{2} xy=
2
1
x
Lets take PC² = PB²
Where P = (x,y) , C = (3,0) and B = (-5,0)
(x - 3)² + (y - 0)² = (x + 5)² + (y - 0)²
x² + 9 - 6x + y² = x² + 25 + 10x + y²
Cancel x² and y² on both sides of the above equation
9 - 6x = 25 + 10x
16 x = 16
x = 1
Substitute x = 1 in y = \frac{1}{2} xy=
2
1
x
Then y = \frac{1}{2}y=
2
1
P (x,y) = (1, \frac{1}{2}1,
2
1
)
Therefore the co-ordinates of Circumcenter of a triangle ABC with vertices A(-3,-4), B(-5,0) and C(3,0) is (1, \frac{1}{2}1,
2
1
)
Answer:
The co-ordinates of the circumcenter of triangle ABC is P (1, \frac{1}{2}1,
2
1
)
Step-by-step explanation:
Given Data
The vertices of given triangle are A (-3,-4), B (-5,0) and C (3,0)
To find - the circumcentre P(x,y) of the triangle ABC.
The point of intersection of all the three perpendicular bisectors of the sides of triangle is called as Circumcenter of a triangle.
PA² = PB² = PC²
Lets take PA² = PB²
Where P = (x,y) , A = (-3,-4) and B = (-5,0)
(x + 3)² + (y +4)² = (x +5)² + (y - 0)²
x² + 9 + 6x + y² + 16 + 8y = x² + 25 + 10x + y²
Cancel x² and y² on both sides of the above equation
9 + 6x + 16 + 8y = 25 + 10x
25 + 8y = 25 + 4x
8y = 4x
y = \frac{1}{2} xy=
2
1
x
Lets take PC² = PB²
Where P = (x,y) , C = (3,0) and B = (-5,0)
(x - 3)² + (y - 0)² = (x + 5)² + (y - 0)²
x² + 9 - 6x + y² = x² + 25 + 10x + y²
Cancel x² and y² on both sides of the above equation
9 - 6x = 25 + 10x
16 x = 16
x = 1
Substitute x = 1 in y = \frac{1}{2} xy=
2
1
x
Then y = \frac{1}{2}y=
2
1
P (x,y) = (1, \frac{1}{2}1,
2
1
)
Therefore the co-ordinates of Circumcenter of a triangle ABC with vertices A(-3,-4), B(-5,0) and C(3,0) is (1, \frac{1}{2}1,
2
1
)
Step-by-step explanation:
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