Question

Points $A(\sqrt{2},4),B(6,-\sqrt{3})$ and $C$ are collinear. If $B$ is the midpoint of line segment $AC$, approximately calculate the $(x,y)$ coordinates of point $C$.

• A. $(3.71,1.13)$
• B. $(3.71,5.73)$
• C. $(7.41,-7.46)$
• D. $(10.59,-7.46)$

Answer 1

The correct option is D $(10.59,-7.46)$

Given points $A$ $(\sqrt{2},4)$, $B$ $(6,-\sqrt{3})$ and $C$ are col-linear if $B$ the mid point $AC$.

if (A) $(3.71,1.13)$ is point $C$, then point $(x,y)=$ $\frac{\sqrt{2}+3.71}{2},\frac{4+1.13}{2}$

$\Rightarrow 2.56,2.56$ this not point $B$

if (B) $(3.71,5.73)$ is point $C$, then point $(x,y)=$ $\frac{\sqrt{2}+3.71}{2},\frac{4+5.73}{2}$

$\Rightarrow 2.56,4.86$ this not point $B$

if (C) $(7.41,-7.46)$ is point $C$, then point $(x,y)=$ $\frac{\sqrt{2}+7.41}{2},\frac{4-7.46}{2}$

$\Rightarrow 4.41,-1.73$ this not point $B$

if (D) $(10.59,-7.46)$ is point $C$, then point $(x,y)=$ $\frac{\sqrt{2}+10.59}{2},\frac{4-7.46}{2}$

$\Rightarrow 6,1.73$ this the point $B(6,1.73)$

if (E) $(10.59,5.73)$ is point $C$, then point $(x,y)=$ $\frac{\sqrt{2}+10.59}{2},\frac{4+5.73}{2}$

$\Rightarrow 6,4.86$ this not point $B$.

So (D) $(10.59,-7.46)$ is mid point $B$ of line $AC$.