$A,B,C$ are three collinear points. The coordinates of $A$ and $B$ are $(3,4)$ and $(7,7)$ respectively and $AC=10$ units. Find the co-ordinates of $C$.

Let the given points are $A(3,4), B(7,7)$ and $c(x, y)$

Distance between any two points is given by

$d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$

Using the above formula,

$\begin{array}{l} A B=\sqrt{(7-3)^{2}+(7-4)^{2}} \\ A B=\sqrt{4^{2}+3^{2}} \\ A B=\sqrt{16+9} \\ A B=\sqrt{25} \\ A B=5 \text { units } \end{array}$

As

$\begin{aligned} A B+B C & =A C \\ 5+B C & =A C \\ 5+B C & =10 \\ B C & =10-5 \\ B C & =5 \text { units } \end{aligned}$

$\therefore A B=B C=5$ units

i.e, $B$ devides $A C$ in the ratio $1: 1 ;$ in, $B$ is Mid-point of AC. Using Mid-point formula,

$\begin{array}{l} \therefore B(7,7)=\frac{3+x}{2}, \frac{4+y}{2} \\ \Rightarrow \quad \frac{3+x}{2}=7 \quad, \quad \frac{4+y}{2}=7 \\ \Rightarrow \quad 3+x=7 \times 2 \quad, \quad 4+y=7 \times 2 \\ \text { (2) Scanned =vith } 4-3=11, \quad y=14-4=10 \\ \text { cangsianctis af } C=(11,10) \\ \end{array}$