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

$A=(3,4),B=(7,7),AC=10$

Let C = (a, b)

G.T A, B, C are collinear
$\Rightarrow \text{ area of }\mathrm{△}ABC=0$

$\begin{array}{l}\Rightarrow \frac{1}{2}\left|\begin{array}{ccl}3& 4& 1\\ 7& 7& 1\\ a& b& 1\end{array}\right|=0\\ \Rightarrow \left|3\right(7-b)-4(7-a)+1(7b-7a\left)\right|=0\end{array}$

$\begin{array}{l}\Rightarrow |21-3\mathrm{b}-28+4\mathrm{a}+7\mathrm{b}-7\mathrm{a}|=0\\ \Rightarrow |4\mathrm{b}-3\mathrm{a}-7|=0\\ 3\mathrm{a}-4\mathrm{b}+7=0\\ \Rightarrow 4\mathrm{b}=3\mathrm{a}+7\\ 4\mathrm{b}-16=3\mathrm{a}-9\\ \mathrm{b}-4=\frac{3}{4}(\mathrm{a}-3)\cdots \cdots \left(1\right)\\ \text{ now, }\mathrm{AC}=10\\ {\mathrm{AC}}^2=100\\ (\mathrm{a}-3{)}^2+(\mathrm{b}-4{)}^2=100\cdots \cdots \left(2\right)\\ \text{ From }\left(1\right)\mathrm{\&}\left(2\right)\\ (\mathrm{a}-3{)}^2+\frac{9}{16}(\mathrm{a}-3{)}^2=100\\ \frac{25}{16}(\mathrm{a}-3{)}^2=100\\ (\mathrm{a}-3{)}^2=8^2\\ \mathrm{a}-3=8,-8\\ \mathrm{a}=11,-5\\ \left(1\right)\Rightarrow \mathrm{b}=10,-2\\ \mathrm{C}=(11,10)\text{ or }(-5,-2)\\ \left(\text{ or }\right)\\ \mathrm{AB}=5,\mathrm{AC}=10\end{array}$

$\left(\frac{a+3}{2}, \frac{b+4}{2}\right)=\left(7,7\right)$

(a,b) = (11, 10)

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