Question

If the points $\left(a_{1}, b_{1}\right)$, $\left(a_{2}, b_{2}\right)$ and $\left(a_{1} + a_{2}, b_{1} + b_{2}\right)$ are collinear, then

• A. $a_{1}b_{2}=a_{2}b_{1}$
• B. $a_{1}+ a_{2} = b_{1} + b_{2}$
• C. $a_{2}b_{2}=a_{1}b_{1}$
• D. $a_{1}+b_{1}=a_{2}+b_{2}$

Answer 1

Answer: (A)

$a_{1}b_{2}=a_{2}b_{1}$

Answer 2

The given points are collinear. $\therefore\quad\frac{1}{2}\left|\begin{matrix}a_{1}&b_{1}&1\\\ a_{2}&b_{2}&1\\\ a_{1}+a_{2}&b_{1}+b_{2}&1\end{matrix}\right|=0$ Applying $R_{2}\rightarrow R_{2}- R_{1}, R_{3}\rightarrow R_{3}- R_{1}$, we get $\left|\begin{matrix}a_{1}&b_{1}&1\\\ a_{2}-a_{1}&b_{2}-b_{1}&0\\\ a_{2}&b_{2}&0\end{matrix}\right|=0$ Expanding along $C_{3}$, we get $b_{2}\left(a_{2}-a_{1}\right)-a_{2}\left(b_{2}-b_{1}\right)=0$ $\Rightarrow\quad-a_{1}b_{2}+a_{2}b_{1}=0 \Rightarrow a_{1}b_{2}=a_{2}b_{1}$