Question

If ${\mathrm{t}}_1,{\mathrm{t}}_2,{\mathrm{t}}_3$ are distinct and the points $\left({\mathrm{t}}_1,2{\mathrm{at}}_1+{{\mathrm{at}}_1}^3\right),\left({\mathrm{t}}_2,2{\mathrm{at}}_2+{{\mathrm{at}}_2}^3\right),\left({\mathrm{t}}_3,2{\mathrm{at}}_3+{{\mathrm{at}}_3}^3\right)$ are collinear then find the value of ${\mathrm{t}}_1+\hspace{0.17em}{\mathrm{t}}_2+{\mathrm{t}}_3$ .

Answer 1

Step 1: Apply condition for three lines to be concurrent in determinant form

$$\begin{gathered} A\equiv \left({\mathrm{t}}_1,2{\mathrm{at}}_1+{{\mathrm{at}}_1}^3\right) \\ B\equiv \left({\mathrm{t}}_2,2{\mathrm{at}}_2+{{\mathrm{at}}_2}^3\right) \\ C\equiv \left({\mathrm{t}}_3,2{\mathrm{at}}_3+{{\mathrm{at}}_3}^3\right) \end{gathered}$$

Given $\mathrm{A},\mathrm{B},\mathrm{C}$ are collinear then by the condition for three points to be collinear in determinant form,

$\left|\begin{array}{ccc}1& {\mathrm{t}}_1& 2a{\mathrm{t}}_1+{{\mathrm{at}}_1}^3\\ 1& {\mathrm{t}}_2& 2a{\mathrm{t}}_2+{{\mathrm{at}}_2}^3\\ 1& {\mathrm{t}}_3& 2a{\mathrm{t}}_3+{{\mathrm{at}}_3}^3\end{array}\right|=0$

A determinant can be split as the sum of two if both determinants have either same two rows or same two columns.

$\begin{array}{ccc}\Rightarrow & \left|\begin{array}{ccc}1& {\mathrm{t}}_1& 2a{\mathrm{t}}_1\\ 1& {\mathrm{t}}_2& 2a{\mathrm{t}}_2\\ 1& {\mathrm{t}}_3& 2a{\mathrm{t}}_3\end{array}\right|+\left|\begin{array}{ccc}1& {\mathrm{t}}_1& {{\mathrm{at}}_1}^3\\ 1& {\mathrm{t}}_2& {{\mathrm{at}}_2}^3\\ 1& {\mathrm{t}}_3& {{\mathrm{at}}_3}^3\end{array}\right|=0& \\ \Rightarrow & 2a\left|\begin{array}{ccc}1& {\mathrm{t}}_1& {\mathrm{t}}_1\\ 1& {\mathrm{t}}_2& {\mathrm{t}}_2\\ 1& {\mathrm{t}}_3& {\mathrm{t}}_3\end{array}\right|+a\left|\begin{array}{ccc}1& {\mathrm{t}}_1& {t_1}^3\\ 1& {\mathrm{t}}_2& {t_2}^3\\ 1& {\mathrm{t}}_3& {t_3}^3\end{array}\right|=0& \\ \Rightarrow & 0+a\left|\begin{array}{ccc}1& {\mathrm{t}}_1& {t_1}^3\\ 1& {\mathrm{t}}_2& {t_2}^3\\ 1& {\mathrm{t}}_3& {t_3}^3\end{array}\right|=0& \left[{\because \det \left(A\right)=0\mathrm{if}\mathrm{C}}_1={\mathrm{C}}_2\mathrm{or}{\mathrm{C}}_2={\mathrm{C}}_3\mathrm{or}{\mathrm{C}}_1={\mathrm{C}}_3\right]\\ \Rightarrow & \left|\begin{array}{ccc}1& {\mathrm{t}}_1& {t_1}^3\\ 1& {\mathrm{t}}_2& {t_2}^3\\ 1& {\mathrm{t}}_3& {t_3}^3\end{array}\right|=0& \end{array}$

[Note: Value of the determinant is $0$ if it has two or more identical rows or columns.]

Step 2: Apply row transformations

Apply row transformations ${\mathrm{R}}_2\to {\mathrm{R}}_2-{\mathrm{R}}_1$ and ${\mathrm{R}}_3\to {\mathrm{R}}_3-{\mathrm{R}}_1$.

$\begin{array}{cc}\Rightarrow & \left|\begin{array}{ccc}1& {\mathrm{t}}_1& {t_1}^3\\ 0& {\mathrm{t}}_2-{\mathrm{t}}_1& {t_2}^3-{t_1}^3\\ 0& {\mathrm{t}}_3-{\mathrm{t}}_1& {t_3}^3-{t_1}^3\end{array}\right|=0\\ \Rightarrow & \left({\mathrm{t}}_2-{\mathrm{t}}_1\right)\left({\mathrm{t}}_3-{\mathrm{t}}_1\right)\left|\begin{array}{ccc}1& {\mathrm{t}}_1& {t_1}^3\\ 0& 1& {{\mathrm{t}}_2}^2+{\mathrm{t}}_2{\mathrm{t}}_1+{{\mathrm{t}}_1}^2\\ 0& 1& {{\mathrm{t}}_3}^2+{\mathrm{t}}_3{\mathrm{t}}_1+{{\mathrm{t}}_1}^2\end{array}\right|=0\\ \Rightarrow & \left({\mathrm{t}}_2-{\mathrm{t}}_1\right)\left({\mathrm{t}}_3-{\mathrm{t}}_1\right)\left({{\mathrm{t}}_2}^2+{\mathrm{t}}_2{\mathrm{t}}_1+{{\mathrm{t}}_1}^2-{{\mathrm{t}}_3}^2-{\mathrm{t}}_3{\mathrm{t}}_1-{{\mathrm{t}}_1}^2\right)=0\\ \Rightarrow & \left({\mathrm{t}}_2-{\mathrm{t}}_1\right)\left({\mathrm{t}}_3-{\mathrm{t}}_1\right)\left({\mathrm{t}}_3-{\mathrm{t}}_2\right)\left({\mathrm{t}}_2+{\mathrm{t}}_1+{\mathrm{t}}_3\right)=0\end{array}$

Since ${\mathrm{t}}_1,{\mathrm{t}}_2,{\mathrm{t}}_3$ are distinct so ${\mathrm{t}}_1\ne {\mathrm{t}}_2\ne {\mathrm{t}}_3$

$\Rightarrow {\mathrm{t}}_1+{\mathrm{t}}_2+{\mathrm{t}}_3=0$

Hence, value of ${\mathrm{t}}_1+\hspace{0.17em}{\mathrm{t}}_2+{\mathrm{t}}_3$ is $0$.