Question

Check whether (7,-3) , (6,4) and (-4,2) are collinear.​

Answer 1

       $\underline{\bold{Solution:}}$

       $\text{We have to check whether }A(7,-3),B(6,4)\text{ and }C(-4,2) \text{ are collinear}$

       $\text{If }A,B,C \text{ are collinear then }Ar\triangle{ABC} = 0$

       $\text{Applying area of triangle formula for coordinates:}$

$\implies \dfrac12\{x_1(y_2-y_3)+x_2(y_3-y_1) + x_3(y_1-y_2)\}$

       $\text{In our case:}$

       $(x_1,y_1) = (7,-3)$

       $(x_2,y_2) = (6,4)$

       $(x_3,y_3) = (-4,2)$

$\implies Ar\triangle{ABC} = \dfrac12[7(4-2) + 6\{2-(-3)\} - 4(-3-4)]$

$\implies Ar\triangle{ABC} = \dfrac12\{7(4-2) + 6(2+3) - 4(-3-4)\}$

$\implies Ar\triangle{ABC} = \dfrac12\{7(2) + 6(5) - 4(-7)\}$

$\implies Ar\triangle{ABC} = \dfrac12(14 + 30 +28)$

$\implies Ar\triangle{ABC} = \dfrac12(72)$

$\implies Ar\triangle{ABC} = 36 \text{ sq. units}$

$\implies Ar\triangle{ABC}\neq 0$

 $\therefore\text{ }\text{ }\boxed{\text{The points }(7,-3),(6,4)\text{ and }(-4,2) \text{ are non-collinear}}$