Find the circumcentre and the area of the triangle with vertices $(7,9), (10,8)$ and $(12,10)$.

(N/A) Let the vertices be $A(7,9), B(10,8)$,and $C(12,10)$.
$1$. Area of the triangle: Using the formula $\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$,we get $\text{Area} = \frac{1}{2} |7(8 - 10) + 10(10 - 9) + 12(9 - 8)| = \frac{1}{2} |-14 + 10 + 12| = \frac{1}{2} |8| = 4 \text{ sq units}$.
$2$. Circumcentre $(h, k)$: The circumcentre is equidistant from the vertices. Thus,$(h-7)^2 + (k-9)^2 = (h-10)^2 + (k-8)^2 = (h-12)^2 + (k-10)^2$.
Solving $(h-7)^2 + (k-9)^2 = (h-10)^2 + (k-8)^2$ gives $6h - 2k = 32$ or $3h - k = 16$.
Solving $(h-10)^2 + (k-8)^2 = (h-12)^2 + (k-10)^2$ gives $4h + 4k = 84$ or $h + k = 21$.
Adding the two equations: $4h = 37 \implies h = \frac{37}{4}$.
Substituting $h$: $\frac{37}{4} + k = 21 \implies k = 21 - \frac{37}{4} = \frac{47}{4}$.
Thus,the circumcentre is $(\frac{37}{4}, \frac{47}{4})$ and the area is $4$.