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
Verified672.3k+ views
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.
Recently Updated Pages
If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE
A car covers the first half distance between two places class 11 physics CBSE
The resultant of two vectors overrightarrow P and overrightarrow class 11 physics CBSE
Find the value of cos 135 class 11 maths CBSE
A mass M is held in place by an applied force F and class 11 physics CBSE
A solution of glucose in water is labelled as 10 dfracwv class 11 chemistry CBSE
Trending doubts
Find the value of the expression given below sin 30circ class 11 maths CBSE
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE
Draw a diagram of nephron and explain its structur class 11 biology CBSE
10 examples of friction in our daily life
Proton was discovered by A Thomson B Rutherford C Chadwick class 11 chemistry CBSE
Bond order ofO2 O2+ O2 and O22 is in order A O2 langle class 11 chemistry CBSE