Question

Herewith I attached the question of maths from coordinate geometry .
# ONLY 16 QUESTION.
# BRAINLIEST
Pls do answer quickly

Answer 1

(16). Let the given coordinates be A(0,0), B(4,0) and C(0,6), D(x,y).

Let the circumcentre of a triangle be (x,y).

(i)

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

$=> \sqrt{x^2 + y^2}$

(ii)

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

$=> \sqrt{x^2 + 16 - 8x + y^2}$

(iii)

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

$=> \sqrt{x^2 + y^2 + 36 - 12y}$

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(1) = (2)

$=> \sqrt{x^2 + y^2} = \sqrt{x^2 + 16 - 8x + y^2}$

On squaring both sides, we get

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

=> 16 = 8x

=> x = 2.    ----- (1)

-------------------------------------------------------------------------------------------------------------

(2) = (3)

$=> \sqrt{x^2 + 16 - 8x + y^2} = \sqrt{x^2 + y^2 + 36 - 12y}$

$=> x^2 - 8x + 16 + y^2 - x^2 - y^2 - 36 + 12y = 0$

$=> 8x - 16 + 36 - 12y = 0$

$=> 8x + 20 = 12y$

$=> 2x + 5 = 3y$

$=> y = \frac{2x + 5}{3}$  ------- (2)

--------------------------------------------------------------------------------------------------------------Substitute (1) in (2) , we get

$=> y = \frac{2(2) + 5}{3}$

$=> y = \frac{9}{3}$

$=> y = 3$

Therefore, the coordinates are (2,3) - Option (C).

Hope this helps!