Question

A(-1,1), B(0,-4), C(-1-5) are the vertices of ∆ABC. find the coordinates of a point in the plane of the triangle which is equidistant from all the three vertices of the triangle​

Answer 1

Answer:

Hope it helps!! Mark this answer as brainliest if u found it useful..

Step-by-step explanation:

Basically, the vertices of circumcenter has been asked .

Now , suppose the vertices are (x,y)

So, (x+1)^2+(y-1)^2 = x^2 + (y+4)^2

x^2 + 1 + 2x + y^2 +1-2y = x^2 + y^2 +16 + 8y

or, 2x-14=10y .....(1)

Similarly, we can find other distances.

x^2 + (y+4)^2 = (x+1)^2+(y+5)^2

or, 16+8y = 1+2x + 25 + 10y

-2y -2x =10 .....(2)

Now add (1) & (2), we have :

-2y-14 = 10y +10

12y =-24

y = -2

So, from (1),

2x-14 =10y

2x -14 = 10*(-2)

2x = -20+14 = -6

So x = -3