Question

A(-3,-4) , B(-5,0) and C(3,0) are the vertices of ∆ ABC . Find the co-ordinates of the circumcentre of ∆ ABC. ​

Answer 1

The co-ordinates of the circumcenter of triangle ABC is P ($1, \frac{1}{2}$ )

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

$y = \frac{1}{2} x$

• 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 $y = \frac{1}{2} x$

Then $y = \frac{1}{2}$

P (x,y) =  ($1, \frac{1}{2}$ )

Therefore the co-ordinates of Circumcenter of a triangle ABC with vertices A(-3,-4), B(-5,0) and C(3,0) is ($1, \frac{1}{2}$ )

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