Question

find the co-ordinate of the circumcentre of a triangle whose vertices are (-3,1) (0,-2) and (1,3) also find its radius correct ans will mark as brainliest​

Answer 1

Answer:

cordinate = (-1/3,2/3) and radius will be √137/3

Step-by-step explanation:

use pA= pB=pC

we know that circumcentre of the traingle is equidistant from vertices of traingle

so

pA^2=pB^2

(x+3)^2+(y-1)^2=(x-0)^2+(y+2)^2

simliary

pB^2=pC^2

we will get the answer

Answer 2

Gɪᴠᴇɴ :-

• vertices of ∆ are (-3,1) (0,-2) and (1,3) .

ᴛᴏ ꜰɪɴᴅ :-

• co-ordinate of the circumcentre and its Radius ?

Sᴏʟᴜᴛɪᴏɴ :-

Let us Assume That, (X, Y) be the coordinates of the circumcenter.

By circumcenter properties, the Distance of (X, Y) from each vertex of a triangle would be the same.

Assume that D1 be the distance between the vertex (x1, y1) and the circumcenter(X, Y).

D2 = Distance b/w (x2,y2) & the circumcenter(X, Y).

D3 = Distance b/w (x3,y3) & the circumcenter(X, Y).

Then By Distance formula we get :-

→ D1 = √[(X - x1)²+(Y - y1)²]

→ D2 = √[(X - x2)²+(Y - y2)²]

→ D3 = √[(X - x3)²+(Y - y3)²]

Now, since D1 = D2 and D2 = D3, we get :-

→ (X − x1)² + (Y − y1)² = (X - x2)²+(Y - y2)²

And,

→ (X - x2)²+(Y - y2)² = (X - x3)²+(Y - y3)²

_______________

Now, we have given That,

→ (x1 , y1) = (-3, 1)

→ (x2 , y2) = (0, -2)

→ (x3 , y3) = (1, 3)

Putting These values now , we get :-

→ (x - (-3))² + (y - 1)² = (x - 0)² + (y - (-2))²

→ (x + 3)² + (y - 1)² = x² + (y + 2)²

→ x² + 6x + 9 + y² - 2y + 1 = x² + y² + 4y + 4

→ 6x + 9 + 1 - 2y - 4y - 4 = 0

→ 6x - 6y + 6 = 0

→ 6(x - y) = -6

→ (x - y) = (-1) --------- Equation (1).

Similarly,

→ (x - 0)² + (y - (-2))² = (x - 1)² + (y - 3)²

→ x² + y² + 4y + 4 = x² - 2x + 1 + y² - 6y + 9

→ 2x + 4y + 6y = 1 + 9 - 4

→ 2x + 10y = 6

→ 2(x + 5y) = 6

→ x + 5y = 3 --------------- Equation (2).

_________________

Now,

Subtracting Equation (1) From Equation (2), we get,

→ (x + 5y) - (x - y) = 3 - (-1)

→ x - x + 5y + y = 3 + 1

→ 6y = 4

→ y = (4/6)

→ y = (2/3)

Putting Value of y in Equation (1) now,

→ x - (2/3) = (-1)

→ x = (-1) + (2/3)

→ x = ( -3 + 2 ) / 3

→ x = (-1/3).

Hence, co-ordinate of the circumcentre of Required ∆ will be {(-1/3) , (2/3)} .

________________________

Now, We know That, Circumcentre will be Equidistant from all Three Vertices . And That Distance is Equal to Radius of Circumcircle .

Hence, we can say That :-

→ Distance b/w (-3,1) & (2/3, -1/3) = (Radius)

→ [(-3) - (2/3)]² + [1 - (-1/3)]² = (Radius)²

→ [ (-9 - 2) / 3 ]² + [ (3 + 1)/3 ]² = (Radius)²

→ (-11/3)² + (4/3)² = (Radius)²

→ (121/9) + (16/9) = (Radius)²

→ (137/9) = (Radius)²

→ Radius = (√137/3) (Ans.)

Hence, Radius of Circumcircle will be (√137/3) .