Question

Find the coordinates of the circumcentre of a triangle whose vertices are A(4,6),B(0,4), and C(6,2).Also find its circumcentre.

Answer 1

$\pink{\underline{\huge{Solution}}}$

Let the coordinates of the circumcentre of the triangle be P(x, y).Circumcentre of a triangle is equidistant from each of the vertices.

So PA = PB = PC

statement : circumcenter is equidistant from the vertices of triangle ABC

Now

$\sqrt{(x-4)^{2}+(y-6)^{2}} = \sqrt{(x-0)^{2} + (y-4)^{2}} = \sqrt{(x-6)^{2}+(y-2)^{2}}$

Now take ,

$\sqrt{(x-4)^{2}+(y-6)^{2}} = \sqrt{(x-0)^{2} + (y-4)^{2}}\\\implies x^{2}+16-8x+y^{2}+36-12x = x^{2}+y^{2}+16-8x\\\implies 16+36-12x=16\\\implies 36=12x\\\implies x = 3$

$\sqrt{(x-0)^{2} + (y-4)^{2}} = \sqrt{(x-6)^{2}+(y-2)^{2}}\\\implies x^{2}+y^{2}+16-8x = x^{2}+36-12x+y^{2}+4-4y\\\implies 16-8x = 36-12x+4-4y\\\implies 16-8(3)=36-12(3)+4-4y\\\implies -8 = 4-4y\\\implies -8-4 =-4y\\\implies -12 = -4y \\\implies y = 3$

Now , circumcenter of ∆ABC = P(3,3)

$\boxed{\orange{Answer=P(3,3)}}$

Answer 2

Answer:

Co-ordinates are (3,3)

Step-by-step explanation:

Solution is as follows

Hope it will help you