Question

A line segment $CD$ has one endpoint $C\left( 6,5 \right)$ and midpoint $M\left( 4,2 \right)$ . How do you determine point $D$ ?

Answer 1

For answering this question we have been asked to find the point $D$ using the given information that there is a line segment $CD$ one of its endpoint is $C\left( 6,5 \right)$ and its midpoint is $M\left( 4,2 \right)$ . We will use the formulae for finding distance between two points $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ from the basic concepts is given as $\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}$ .

Complete answer:
Now considering from the question we have been given the information stated as “A line segment $CD$ has one endpoint $C\left( 6,5 \right)$ and midpoint $M\left( 4,2 \right)$” we need to find the point $D$ .
We know that the formulae for finding distance between two points $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ from the basic concepts is given as $\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}$ .
From the basic concept we know that a midpoint of a line segment is that point which divides the line segment into two equal halves having the same length.

So we can say that the distance between $C$ and $M$ we will be equal to the distance between $D$ and $M$ .
For the time being let us assume $D$ as $\left( x,y \right)$ .
The distance between $C\left( 6,5 \right)$ and $M\left( 4,2 \right)$ will be $\Rightarrow \sqrt{{{\left( 6-4 \right)}^{2}}+{{\left( 5-2 \right)}^{2}}}$
Now by performing further simplifications we will have the distance as
$\begin{aligned} & \Rightarrow \sqrt{{{2}^{2}}+{{3}^{2}}}=\sqrt{4+9} \\\ & \Rightarrow \sqrt{13} \\\ \end{aligned}$ .
So we can say that the distance between $D\left( x,y \right)$ and $M\left( 4,2 \right)$ will be $\Rightarrow \sqrt{13}$ .
After this we will get the equation $\sqrt{{{\left( 4-x \right)}^{2}}+{{\left( 2-y \right)}^{2}}}=\sqrt{13}$ .
After performing basic arithmetic simplifications we will have $\Rightarrow {{\left( 4-x \right)}^{2}}+{{\left( 2-y \right)}^{2}}=13$ .
We know that $13$ can be written as the sum of $4$ and $9$ .
So now we will have $\Rightarrow {{\left( 4-x \right)}^{2}}+{{\left( 2-y \right)}^{2}}=4+9$ .
So now we can say that $4-x=2\Rightarrow x=2$ and $2-y=3\Rightarrow y=-1$ .
Therefore we can conclude that for a line segment $CD$ having one endpoint $C\left( 6,5 \right)$ and midpoint $M\left( 4,2 \right)$then the point $D$ will be $\left( 2,-1 \right)$ .

Note:
We should be sure with our calculations while solving this question and the transformations and basic arithmetic simplifications we make. This question can be also solved by using the formulae for finding the midpoint of a line segment having the end points $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ will be given as $\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right)$ . By applying this formulae in this question we will have the equation $\left( 4,2 \right)=\left( \dfrac{x+6}{2},\dfrac{y+5}{2} \right)$ after simplifying it we will obtain $x=2$ and $y=-1$ .