If A,B,C are collinear points such that $\mathrm{A}=(3,4),\mathrm{B}=(7,7)$ and AC = 10, then C = 

$\mathrm{A}=(3,4),\mathrm{B}=(7,7),\mathrm{AC}=10$
Let C= (a,b)
G.T A,B,C are collinear
$\begin{array}{l}\Rightarrow \mathrm{area} \mathrm{of} \mathrm{\Delta ABC}=0\\ \Rightarrow \frac{1}{2}\left|\begin{array}{lll}3& 4& 1\\ 7& 7& 1\\ \mathrm{a}& \mathrm{b}& 1\end{array}\right|=0\\ \Rightarrow \left|3\right(7-\mathrm{b})-4(7-\mathrm{a})+1(7\mathrm{b}-7\mathrm{a}\left)\right|=0\\ \Rightarrow |21-3\mathrm{b}-28+4\mathrm{a}+7\mathrm{b}-7\mathrm{a}|=0\\ \Rightarrow |4\mathrm{b}-3\mathrm{a}-7|=0\\ 3\mathrm{a}-4\mathrm{b}+7=0\\ \Rightarrow 4\mathrm{b}=3\mathrm{a}+7\\ 4\mathrm{b}-16=3\mathrm{a}-9\\ \mathrm{b}-4=\frac{3}{4}(\mathrm{a}-3)\cdots \cdots \left(1\right)\\ \mathrm{now}, \mathrm{AC}=10\\ {\mathrm{AC}}^2=100\\ (\mathrm{a}-3{)}^2+(\mathrm{b}-4{)}^2=100\cdots \cdots \left(2\right)\end{array}$
From (1) & (2)
$\begin{array}{l}(\mathrm{a}-3{)}^2+\frac{9}{16}(\mathrm{a}-3{)}^2=100\\ \frac{25}{16}(\mathrm{a}-3{)}^2=100\\ (\mathrm{a}-3{)}^2=8^2\\ \mathrm{a}-3=8,-8\\ \mathrm{a}=11,-5\\ \left(1\right)\Rightarrow \mathrm{b}=10,-2\\ \mathrm{C}=(11,10) \mathrm{or} (-5,-2)\end{array}$
(Or)
AB = 5, AC = 10
 

$\begin{array}{l}\left(\frac{\mathrm{a}+3}{2},\frac{\mathrm{b}+4}{2}\right)=(7,7)\\ (\mathrm{a},\mathrm{b})=(11,10)\end{array}$
 

$\begin{array}{l}\mathrm{CA}:\mathrm{AB}=2:1\\ \mathrm{A}(3,4)=\left(\frac{2\left(7\right)+\mathrm{a}}{2+1},\frac{2\left(7\right)+1\left(\mathrm{b}\right)}{2+1}\right)\\ (3,4)=\left(\frac{14+\mathrm{a}}{3},\frac{14+\mathrm{b}}{3}\right)\\ (\mathrm{a},\mathrm{b})=(-5,-2)\end{array}$