Question

If $A,B,C$ are collinear points such that $A(3,4),C(11,10)$ and $AB=2.5$ then point $B$ is

• A. $(5,\frac{11}{2})$
• B. $(\frac{5}{2},11)$
• C. $(5,5)$
• D. $(5,6)$

Answer 1

The correct option is A $(5,\frac{11}{2})$

Given that the points $A,B,C$ are collinear such that $A(3,4),C(11,10)$ and $AB=2.5$,

Consider the line joining $A$ and $C$,

Slope of that line is $\tan \alpha =\frac{10-4}{11-3}=\frac{3}{4}$,

$⟹\cos \alpha =\frac{4}{5}$ and $\sin \alpha =\frac{3}{5}$,

We know that a point in a line at a distance $r$ from that point is $y=y_1+r\sin \alpha$ and $x=x_1+r\cos \alpha$,

Where $\tan \alpha$ is the slope of that line

Let the coordinates of the point $B$ be $(x,y)$,

$⟹x=3+2.5\left(0.2\right)=5$ and $y=3+2.5\left(0.6\right)=\frac{11}{2}$,

$\therefore$ The coordinates of $B$ are $(5,\frac{11}{2})$