Determinants — Class 12 Maths Solution

(i) Equation of the line joining (${x_1},{y_1})$ and $({x_2},{y_2})$

is $\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}x&y&1\\{{x_1}}&{{y_1}}&1\\{{x_2}}&{{y_2}}&1\end{array}} \right| = 0$

$\Rightarrow$ $\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}x&y&1\\1&2&1\\3&6&1\end{array}} \right| = 0$

$\Rightarrow$ $x(2 - 6) - y(1 - 3) + 1(6 - 6) = 0$

$\Rightarrow$ $- 4x + 2y = 0 \Rightarrow 2x - y = 0$

Hence, $y = 2x$ is the required line.

(ii) Equation of the line joining $({x_1},{y_1})$ and $({x_2},{y_2})$
is$\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}x&y&1\\{{x_1}}&{{y_1}}&1\\{{x_2}}&{{y_2}}&1\end{array}} \right| = 0$

$\Rightarrow$ $\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}x&y&1\\3&1&1\\9&3&1\end{array}} \right| = 0$

$\Rightarrow$ $x(1 - 3) - y(3 - 9) + 1(9 - 9) = 0 \Rightarrow - 2x + 6y = 0$

$\Rightarrow$ $- x + 3y = 0 \Rightarrow x - 3y = 0$

Hence, $x - 3y = 0$ is the required line.