If the point $\left( {{a}_{1}},{{b}_{1}} \right),\,\left( {{a}_{2}},{{b}_{2}} \right)\,\text{and}\,\left[ \left( {{a}_{1}}+{{a}_{2}} \right),\left( {{b}_{1}}+{{b}_{2}} \right) \right]$ are collinear, then show that ${{a}_{1}}{{b}_{2}}={{a}_{2}}{{b}_{1}}$.
Answer
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Hint:In this question, we will use the distance between two points formula of two-dimensional geometry and apply the condition that the sum of distance between two pairs of points is equal to the third pair of points.
Complete step-by-step answer:
Let $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$be two points in two-dimensional geometry. Then distance, let say $d$, between these two points is given by,
$d=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}\cdots \cdots \left( i \right)$
Now, three points lying in a two-dimensional plane are collinear if and only if the sum of distance between two pairs of points is equal to the third pair of points. That is, if three points are A, B and C are collinear, then, AB + BC = AC.
Let, $A=\left( {{a}_{1}},{{b}_{1}} \right),\,B=\,\left( {{a}_{2}},{{b}_{2}} \right)\,\text{and}\,C=\left[ \left( {{a}_{1}}+{{a}_{2}} \right),\left( {{b}_{1}}+{{b}_{2}} \right) \right]$.
Here, from equation $\left( i \right)$, distance between points $A$ and $B$ is given as,
$AB=\sqrt{{{\left( {{a}_{2}}-{{a}_{1}} \right)}^{2}}+{{\left( {{b}_{2}}-{{b}_{1}} \right)}^{2}}}$
Similarly, distance between points $B$ and $C$ is given as,
$\begin{align}
& BC=\sqrt{{{\left( \left( {{a}_{1}}+{{a}_{2}} \right)-{{a}_{2}} \right)}^{2}}+{{\left( \left( {{b}_{1}}+{{b}_{2}} \right)-{{b}_{2}} \right)}^{2}}} \\
& =\sqrt{{{\left( {{a}_{1}}+{{a}_{2}}-{{a}_{2}} \right)}^{2}}+{{\left( {{b}_{1}}+{{b}_{2}}-{{b}_{2}} \right)}^{2}}} \\
& =\sqrt{{{a}_{1}}^{2}+{{b}_{1}}^{2}} \\
\end{align}$
And, distance between points $A$ and $C$ is given as,
$\begin{align}
& AC=\sqrt{{{\left( \left( {{a}_{1}}+{{a}_{2}} \right)-{{a}_{1}} \right)}^{2}}+{{\left( \left( {{b}_{1}}+{{b}_{2}} \right)-{{b}_{1}} \right)}^{2}}} \\
& =\sqrt{{{\left( {{a}_{1}}+{{a}_{2}}-{{a}_{1}} \right)}^{2}}+{{\left( {{b}_{1}}+{{b}_{2}}-{{b}_{1}} \right)}^{2}}} \\
& =\sqrt{{{a}_{2}}^{2}+{{b}_{2}}^{2}} \\
\end{align}$
Now, since these points are collinear, therefore,
$AB+BC=AC$.
Putting values of $AB,\,BC\,and\,AC$from above, we get,
$\sqrt{{{\left( {{a}_{2}}-{{a}_{1}} \right)}^{2}}+{{\left( {{b}_{2}}-{{b}_{1}} \right)}^{2}}}+\sqrt{{{a}_{1}}^{2}+{{b}_{1}}^{2}}=\sqrt{{{a}_{2}}^{2}+{{b}_{2}}^{2}}$
Subtracting, $\sqrt{{{a}_{1}}^{2}+{{b}_{1}}^{2}}$ from both sides of the equation, we get,
$\sqrt{{{\left( {{a}_{2}}-{{a}_{1}} \right)}^{2}}+{{\left( {{b}_{2}}-{{b}_{1}} \right)}^{2}}}=\sqrt{{{a}_{2}}^{2}+{{b}_{2}}^{2}}-\sqrt{{{a}_{1}}^{2}+{{b}_{1}}^{2}}$
Squaring both sides of the equation, we get,
$\begin{align}
& {{\left[ \sqrt{{{\left( {{a}_{2}}-{{a}_{1}} \right)}^{2}}+{{\left( {{b}_{2}}-{{b}_{1}} \right)}^{2}}} \right]}^{2}}={{\left[ \sqrt{{{a}_{2}}^{2}+{{b}_{2}}^{2}}-\sqrt{{{a}_{1}}^{2}+{{b}_{1}}^{2}} \right]}^{2}} \\
& \Rightarrow {{\left( {{a}_{2}}-{{a}_{1}} \right)}^{2}}+{{\left( {{b}_{2}}-{{b}_{1}} \right)}^{2}}={{\left[ \sqrt{{{a}_{2}}^{2}+{{b}_{2}}^{2}}-\sqrt{{{a}_{1}}^{2}+{{b}_{1}}^{2}} \right]}^{2}} \\
\end{align}$
Applying, $\left( x+y \right)={{x}^{2}}+{{y}^{2}}+2xy$ formula on both sides of equation, we get,
$\begin{align}
& \left( {{a}_{2}}^{2}+{{a}_{1}}^{2}-2{{a}_{1}}{{a}_{2}} \right)+\left( {{b}_{2}}^{2}+{{b}_{1}}^{2}-2{{b}_{1}}{{b}_{2}} \right)={{\left( \sqrt{{{a}_{2}}^{2}+{{b}_{2}}^{2}} \right)}^{2}}+{{\left( \sqrt{{{a}_{1}}^{2}+{{b}_{1}}^{2}} \right)}^{2}}-2\sqrt{{{a}_{2}}^{2}+{{b}_{2}}^{2}}\sqrt{{{a}_{1}}^{2}+{{b}_{1}}^{2}} \\
& \Rightarrow {{a}_{2}}^{2}+{{a}_{1}}^{2}-2{{a}_{1}}{{a}_{2}}+{{b}_{2}}^{2}+{{b}_{1}}^{2}-2{{b}_{1}}{{b}_{2}}={{a}_{2}}^{2}+{{b}_{2}}^{2}+{{a}_{1}}^{2}+{{b}_{1}}^{2}-2\sqrt{\left( {{a}_{2}}^{2}+{{b}_{2}}^{2} \right)\left( {{a}_{1}}^{2}+{{b}_{1}}^{2} \right)} \\
\end{align}$
Subtracting ${{a}_{1}}^{2}+{{a}_{2}}^{2}+{{b}_{1}}^{2}+{{b}_{2}}^{2}$ from both sides of the equation, we get,
$-2{{a}_{1}}{{a}_{2}}-2{{b}_{1}}{{b}_{2}}=-2\sqrt{\left( {{a}_{2}}^{2}+{{b}_{2}}^{2} \right)\left( {{a}_{1}}^{2}+{{b}_{1}}^{2} \right)}$
Dividing $-2$ from both sides of the equation, we get,
${{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}=\sqrt{\left( {{a}_{2}}^{2}+{{b}_{2}}^{2} \right)\left( {{a}_{1}}^{2}+{{b}_{1}}^{2} \right)}$
Again, squaring both sides of the equation, we get,
$\begin{align}
& {{\left( {{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}} \right)}^{2}}={{\left( \sqrt{\left( {{a}_{2}}^{2}+{{b}_{2}}^{2} \right)\left( {{a}_{1}}^{2}+{{b}_{1}}^{2} \right)} \right)}^{2}} \\
& \Rightarrow {{\left( {{a}_{1}}{{a}_{2}} \right)}^{2}}+{{\left( {{b}_{1}}{{b}_{2}} \right)}^{2}}+2{{a}_{1}}{{a}_{2}}{{b}_{1}}{{b}_{2}}=\left( {{a}_{2}}^{2}+{{b}_{2}}^{2} \right)\left( {{a}_{1}}^{2}+{{b}_{1}}^{2} \right) \\
\end{align}$
Applying, distributive law on right side of the equation, we get,
$\begin{align}
& {{a}_{1}}^{2}{{a}_{2}}^{2}+{{b}_{1}}^{2}{{b}_{2}}^{2}+2{{a}_{1}}{{a}_{2}}{{b}_{1}}{{b}_{2}}={{a}_{2}}^{2}{{a}_{1}}^{2}+{{a}_{2}}^{2}{{b}_{1}}^{2}+{{b}_{2}}^{2}{{a}_{1}}^{2}+{{b}_{2}}^{2}{{b}_{1}}^{2} \\
& \Rightarrow {{a}_{1}}^{2}{{a}_{2}}^{2}+{{b}_{1}}^{2}{{b}_{2}}^{2}+2{{a}_{1}}{{a}_{2}}{{b}_{1}}{{b}_{2}}={{a}_{1}}^{2}{{a}_{2}}^{2}+{{b}_{1}}^{2}{{b}_{2}}^{2}+{{b}_{2}}^{2}{{a}_{1}}^{2}+{{a}_{2}}^{2}{{b}_{1}}^{2} \\
\end{align}$
Subtracting ${{a}_{1}}^{2}{{a}_{2}}^{2}+{{b}_{1}}^{2}{{b}_{2}}^{2}$ from both sides of the equation, we have,
$2{{a}_{1}}{{a}_{2}}{{b}_{1}}{{b}_{2}}={{b}_{2}}^{2}{{a}_{1}}^{2}+{{a}_{2}}^{2}{{b}_{1}}^{2}$
Subtracting $2{{a}_{1}}{{a}_{2}}{{b}_{1}}{{b}_{2}}$ to both sides of the equation, we get,
$\begin{align}
& {{b}_{2}}^{2}{{a}_{1}}^{2}+{{a}_{2}}^{2}{{b}_{1}}^{2}-2{{a}_{1}}{{a}_{2}}{{b}_{1}}{{b}_{2}}=0 \\
& \Rightarrow {{\left( {{b}_{2}}{{a}_{1}} \right)}^{2}}+{{\left( {{a}_{2}}{{b}_{1}} \right)}^{2}}-2\left( {{b}_{2}}{{a}_{1}} \right)\left( {{a}_{2}}{{b}_{1}} \right)=0 \\
\end{align}$
Using $\left( x+y \right)={{x}^{2}}+{{y}^{2}}+2xy$, we can write above equation as,
${{\left( {{b}_{2}}{{a}_{1}}-{{a}_{2}}{{b}_{1}} \right)}^{2}}=0$
Taking root on both sides, we get,
${{b}_{2}}{{a}_{1}}-{{a}_{2}}{{b}_{1}}=0$
Adding ${{a}_{2}}{{b}_{1}}$ on both sides of the equation, we get,
$\begin{align}
& {{b}_{2}}{{a}_{1}}={{a}_{2}}{{b}_{1}} \\
& \Rightarrow {{a}_{1}}{{b}_{2}}={{a}_{2}}{{b}_{1}} \\
\end{align}$
Hence, we shown that, if points $\left( {{a}_{1}},{{b}_{1}} \right),\,\left( {{a}_{2}},{{b}_{2}} \right)\,\text{and}\,\left[ \left( {{a}_{1}}+{{a}_{2}} \right),\left( {{b}_{1}}+{{b}_{2}} \right) \right]$ are collinear, then ${{a}_{1}}{{b}_{2}}={{a}_{2}}{{b}_{1}}$.
Note: In this type of questions, if the order of points in which they are arranged is not mentioned, as in this question also, take the order to be the same as in which points are written in the question.
Complete step-by-step answer:
Let $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$be two points in two-dimensional geometry. Then distance, let say $d$, between these two points is given by,
$d=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}\cdots \cdots \left( i \right)$
Now, three points lying in a two-dimensional plane are collinear if and only if the sum of distance between two pairs of points is equal to the third pair of points. That is, if three points are A, B and C are collinear, then, AB + BC = AC.
Let, $A=\left( {{a}_{1}},{{b}_{1}} \right),\,B=\,\left( {{a}_{2}},{{b}_{2}} \right)\,\text{and}\,C=\left[ \left( {{a}_{1}}+{{a}_{2}} \right),\left( {{b}_{1}}+{{b}_{2}} \right) \right]$.
Here, from equation $\left( i \right)$, distance between points $A$ and $B$ is given as,
$AB=\sqrt{{{\left( {{a}_{2}}-{{a}_{1}} \right)}^{2}}+{{\left( {{b}_{2}}-{{b}_{1}} \right)}^{2}}}$
Similarly, distance between points $B$ and $C$ is given as,
$\begin{align}
& BC=\sqrt{{{\left( \left( {{a}_{1}}+{{a}_{2}} \right)-{{a}_{2}} \right)}^{2}}+{{\left( \left( {{b}_{1}}+{{b}_{2}} \right)-{{b}_{2}} \right)}^{2}}} \\
& =\sqrt{{{\left( {{a}_{1}}+{{a}_{2}}-{{a}_{2}} \right)}^{2}}+{{\left( {{b}_{1}}+{{b}_{2}}-{{b}_{2}} \right)}^{2}}} \\
& =\sqrt{{{a}_{1}}^{2}+{{b}_{1}}^{2}} \\
\end{align}$
And, distance between points $A$ and $C$ is given as,
$\begin{align}
& AC=\sqrt{{{\left( \left( {{a}_{1}}+{{a}_{2}} \right)-{{a}_{1}} \right)}^{2}}+{{\left( \left( {{b}_{1}}+{{b}_{2}} \right)-{{b}_{1}} \right)}^{2}}} \\
& =\sqrt{{{\left( {{a}_{1}}+{{a}_{2}}-{{a}_{1}} \right)}^{2}}+{{\left( {{b}_{1}}+{{b}_{2}}-{{b}_{1}} \right)}^{2}}} \\
& =\sqrt{{{a}_{2}}^{2}+{{b}_{2}}^{2}} \\
\end{align}$
Now, since these points are collinear, therefore,
$AB+BC=AC$.
Putting values of $AB,\,BC\,and\,AC$from above, we get,
$\sqrt{{{\left( {{a}_{2}}-{{a}_{1}} \right)}^{2}}+{{\left( {{b}_{2}}-{{b}_{1}} \right)}^{2}}}+\sqrt{{{a}_{1}}^{2}+{{b}_{1}}^{2}}=\sqrt{{{a}_{2}}^{2}+{{b}_{2}}^{2}}$
Subtracting, $\sqrt{{{a}_{1}}^{2}+{{b}_{1}}^{2}}$ from both sides of the equation, we get,
$\sqrt{{{\left( {{a}_{2}}-{{a}_{1}} \right)}^{2}}+{{\left( {{b}_{2}}-{{b}_{1}} \right)}^{2}}}=\sqrt{{{a}_{2}}^{2}+{{b}_{2}}^{2}}-\sqrt{{{a}_{1}}^{2}+{{b}_{1}}^{2}}$
Squaring both sides of the equation, we get,
$\begin{align}
& {{\left[ \sqrt{{{\left( {{a}_{2}}-{{a}_{1}} \right)}^{2}}+{{\left( {{b}_{2}}-{{b}_{1}} \right)}^{2}}} \right]}^{2}}={{\left[ \sqrt{{{a}_{2}}^{2}+{{b}_{2}}^{2}}-\sqrt{{{a}_{1}}^{2}+{{b}_{1}}^{2}} \right]}^{2}} \\
& \Rightarrow {{\left( {{a}_{2}}-{{a}_{1}} \right)}^{2}}+{{\left( {{b}_{2}}-{{b}_{1}} \right)}^{2}}={{\left[ \sqrt{{{a}_{2}}^{2}+{{b}_{2}}^{2}}-\sqrt{{{a}_{1}}^{2}+{{b}_{1}}^{2}} \right]}^{2}} \\
\end{align}$
Applying, $\left( x+y \right)={{x}^{2}}+{{y}^{2}}+2xy$ formula on both sides of equation, we get,
$\begin{align}
& \left( {{a}_{2}}^{2}+{{a}_{1}}^{2}-2{{a}_{1}}{{a}_{2}} \right)+\left( {{b}_{2}}^{2}+{{b}_{1}}^{2}-2{{b}_{1}}{{b}_{2}} \right)={{\left( \sqrt{{{a}_{2}}^{2}+{{b}_{2}}^{2}} \right)}^{2}}+{{\left( \sqrt{{{a}_{1}}^{2}+{{b}_{1}}^{2}} \right)}^{2}}-2\sqrt{{{a}_{2}}^{2}+{{b}_{2}}^{2}}\sqrt{{{a}_{1}}^{2}+{{b}_{1}}^{2}} \\
& \Rightarrow {{a}_{2}}^{2}+{{a}_{1}}^{2}-2{{a}_{1}}{{a}_{2}}+{{b}_{2}}^{2}+{{b}_{1}}^{2}-2{{b}_{1}}{{b}_{2}}={{a}_{2}}^{2}+{{b}_{2}}^{2}+{{a}_{1}}^{2}+{{b}_{1}}^{2}-2\sqrt{\left( {{a}_{2}}^{2}+{{b}_{2}}^{2} \right)\left( {{a}_{1}}^{2}+{{b}_{1}}^{2} \right)} \\
\end{align}$
Subtracting ${{a}_{1}}^{2}+{{a}_{2}}^{2}+{{b}_{1}}^{2}+{{b}_{2}}^{2}$ from both sides of the equation, we get,
$-2{{a}_{1}}{{a}_{2}}-2{{b}_{1}}{{b}_{2}}=-2\sqrt{\left( {{a}_{2}}^{2}+{{b}_{2}}^{2} \right)\left( {{a}_{1}}^{2}+{{b}_{1}}^{2} \right)}$
Dividing $-2$ from both sides of the equation, we get,
${{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}=\sqrt{\left( {{a}_{2}}^{2}+{{b}_{2}}^{2} \right)\left( {{a}_{1}}^{2}+{{b}_{1}}^{2} \right)}$
Again, squaring both sides of the equation, we get,
$\begin{align}
& {{\left( {{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}} \right)}^{2}}={{\left( \sqrt{\left( {{a}_{2}}^{2}+{{b}_{2}}^{2} \right)\left( {{a}_{1}}^{2}+{{b}_{1}}^{2} \right)} \right)}^{2}} \\
& \Rightarrow {{\left( {{a}_{1}}{{a}_{2}} \right)}^{2}}+{{\left( {{b}_{1}}{{b}_{2}} \right)}^{2}}+2{{a}_{1}}{{a}_{2}}{{b}_{1}}{{b}_{2}}=\left( {{a}_{2}}^{2}+{{b}_{2}}^{2} \right)\left( {{a}_{1}}^{2}+{{b}_{1}}^{2} \right) \\
\end{align}$
Applying, distributive law on right side of the equation, we get,
$\begin{align}
& {{a}_{1}}^{2}{{a}_{2}}^{2}+{{b}_{1}}^{2}{{b}_{2}}^{2}+2{{a}_{1}}{{a}_{2}}{{b}_{1}}{{b}_{2}}={{a}_{2}}^{2}{{a}_{1}}^{2}+{{a}_{2}}^{2}{{b}_{1}}^{2}+{{b}_{2}}^{2}{{a}_{1}}^{2}+{{b}_{2}}^{2}{{b}_{1}}^{2} \\
& \Rightarrow {{a}_{1}}^{2}{{a}_{2}}^{2}+{{b}_{1}}^{2}{{b}_{2}}^{2}+2{{a}_{1}}{{a}_{2}}{{b}_{1}}{{b}_{2}}={{a}_{1}}^{2}{{a}_{2}}^{2}+{{b}_{1}}^{2}{{b}_{2}}^{2}+{{b}_{2}}^{2}{{a}_{1}}^{2}+{{a}_{2}}^{2}{{b}_{1}}^{2} \\
\end{align}$
Subtracting ${{a}_{1}}^{2}{{a}_{2}}^{2}+{{b}_{1}}^{2}{{b}_{2}}^{2}$ from both sides of the equation, we have,
$2{{a}_{1}}{{a}_{2}}{{b}_{1}}{{b}_{2}}={{b}_{2}}^{2}{{a}_{1}}^{2}+{{a}_{2}}^{2}{{b}_{1}}^{2}$
Subtracting $2{{a}_{1}}{{a}_{2}}{{b}_{1}}{{b}_{2}}$ to both sides of the equation, we get,
$\begin{align}
& {{b}_{2}}^{2}{{a}_{1}}^{2}+{{a}_{2}}^{2}{{b}_{1}}^{2}-2{{a}_{1}}{{a}_{2}}{{b}_{1}}{{b}_{2}}=0 \\
& \Rightarrow {{\left( {{b}_{2}}{{a}_{1}} \right)}^{2}}+{{\left( {{a}_{2}}{{b}_{1}} \right)}^{2}}-2\left( {{b}_{2}}{{a}_{1}} \right)\left( {{a}_{2}}{{b}_{1}} \right)=0 \\
\end{align}$
Using $\left( x+y \right)={{x}^{2}}+{{y}^{2}}+2xy$, we can write above equation as,
${{\left( {{b}_{2}}{{a}_{1}}-{{a}_{2}}{{b}_{1}} \right)}^{2}}=0$
Taking root on both sides, we get,
${{b}_{2}}{{a}_{1}}-{{a}_{2}}{{b}_{1}}=0$
Adding ${{a}_{2}}{{b}_{1}}$ on both sides of the equation, we get,
$\begin{align}
& {{b}_{2}}{{a}_{1}}={{a}_{2}}{{b}_{1}} \\
& \Rightarrow {{a}_{1}}{{b}_{2}}={{a}_{2}}{{b}_{1}} \\
\end{align}$
Hence, we shown that, if points $\left( {{a}_{1}},{{b}_{1}} \right),\,\left( {{a}_{2}},{{b}_{2}} \right)\,\text{and}\,\left[ \left( {{a}_{1}}+{{a}_{2}} \right),\left( {{b}_{1}}+{{b}_{2}} \right) \right]$ are collinear, then ${{a}_{1}}{{b}_{2}}={{a}_{2}}{{b}_{1}}$.
Note: In this type of questions, if the order of points in which they are arranged is not mentioned, as in this question also, take the order to be the same as in which points are written in the question.
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