Find equation of line joining $(1,2)$ and $(3,6)$ using determinates
Find equation of line joining $(1,2)$ and $(3,6)$ using determinates
- A
$y=2 x$
- B
$y=3 x$
- C
$y=4 x+7$
- D
$y=2 x+9$
Similar Questions
If $\alpha ,\beta \ne 0$ and $f\left( n \right) = {\alpha ^n} + {\beta ^n}$ and $\left| {\begin{array}{*{20}{c}}3&{1 + f\left( 1 \right)}&{1 + f\left( 2 \right)}\\{1 + f\left( 1 \right)}&{1 + f\left( 2 \right)}&{1 + f\left( 3 \right)}\\{1 + f\left( 2 \right)}&{1 + f\left( 3 \right)}&{1 + f\left( 4 \right)}\end{array}} \right|\; = K{\left( {1 - \alpha } \right)^2}$ ${\left( {1 - \beta } \right)^2}{\left( {\alpha - \beta } \right)^2}$ ,then $K=$ . . . . . .
If $\alpha ,\beta \ne 0$ and $f\left( n \right) = {\alpha ^n} + {\beta ^n}$ and $\left| {\begin{array}{*{20}{c}}3&{1 + f\left( 1 \right)}&{1 + f\left( 2 \right)}\\{1 + f\left( 1 \right)}&{1 + f\left( 2 \right)}&{1 + f\left( 3 \right)}\\{1 + f\left( 2 \right)}&{1 + f\left( 3 \right)}&{1 + f\left( 4 \right)}\end{array}} \right|\; = K{\left( {1 - \alpha } \right)^2}$ ${\left( {1 - \beta } \right)^2}{\left( {\alpha - \beta } \right)^2}$ ,then $K=$ . . . . . .
- [JEE MAIN 2014]
If $f\left( x \right) = \left| {\begin{array}{*{20}{c}}
{\sin \left( {x + \alpha } \right)}&{\sin \left( {x + \beta } \right)}&{\sin \left( {x + \gamma } \right)} \\
{\cos \left( {x + \alpha } \right)}&{\cos \left( {x + \beta } \right)}&{\cos \left( {x + \gamma } \right)} \\
{\sin \left( {\alpha + \beta } \right)}&{\sin \left( {\beta + \gamma } \right)}&{\sin \left( {\gamma + \alpha } \right)}
\end{array}} \right|$ and $f(10) = 10$ then $f(\pi)$ is equal to
If $f\left( x \right) = \left| {\begin{array}{*{20}{c}}
{\sin \left( {x + \alpha } \right)}&{\sin \left( {x + \beta } \right)}&{\sin \left( {x + \gamma } \right)} \\
{\cos \left( {x + \alpha } \right)}&{\cos \left( {x + \beta } \right)}&{\cos \left( {x + \gamma } \right)} \\
{\sin \left( {\alpha + \beta } \right)}&{\sin \left( {\beta + \gamma } \right)}&{\sin \left( {\gamma + \alpha } \right)}
\end{array}} \right|$ and $f(10) = 10$ then $f(\pi)$ is equal to
Show that points $A(a, b+c), B(b, c+a), C(c, a+b)$ are collinear
Show that points $A(a, b+c), B(b, c+a), C(c, a+b)$ are collinear
If ${x^a}{y^b} = {e^m},{x^c}{y^d} = {e^n},{\Delta _1} = \left| {\,\begin{array}{*{20}{c}}m&b\\n&d\end{array}\,} \right|\,\,{\Delta _2} = \left| {\,\begin{array}{*{20}{c}}a&m\\c&n\end{array}\,} \right|$ and ${\Delta _3} = \left| {\,\begin{array}{*{20}{c}}a&b\\c&d\end{array}\,} \right|$, then the values of $x$ and $y$ are respectively
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Two fair dice are thrown. The numbers on them are taken as $\lambda$ and $\mu$, and a system of linear equations
$x+y+z=5$ ; $x+2 y+3 z=\mu$ ; $x+3 y+\lambda z=1$
is constructed. If $\mathrm{p}$ is the probability that the system has a unique solution and $\mathrm{q}$ is the probability that the system has no solution, then :
Two fair dice are thrown. The numbers on them are taken as $\lambda$ and $\mu$, and a system of linear equations
$x+y+z=5$ ; $x+2 y+3 z=\mu$ ; $x+3 y+\lambda z=1$
is constructed. If $\mathrm{p}$ is the probability that the system has a unique solution and $\mathrm{q}$ is the probability that the system has no solution, then :
- [JEE MAIN 2021]