Question

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

• A. $(10,11)$
• B. $(-11,10)$
• C. $(11,10)$
• D. $(10,-12)$

Answer 1

The correct option is C $(11,10)$

$AB=\sqrt{\left[\right(7-3{)}^2+(7-4{)}^2]}=\sqrt{[16+9]}=\sqrt{25}=5units$

But,

$AC=AB+BC\Rightarrow BC=AC-AB$

$BC=10-5=5$ units

Therefore B is the midpoint of AC.

Using section formula,

$B(7,7)=\left[\right(x+3\left)/2\right],\left[\right(y+4\left)/2\right]$

i.e., $(x+3)/2=7$ implies $x=11$

and $(y+4)/2=7$ implies $y=10$

therefore $C=(11,10)$