Question

(37R(0, 3), D(2, 1), S(3,-1) determine whether the points are collinear​

Answer 1

Collinear points are those points that lie on the same line.

For three or more co-ordinates to be collinear, the area formed by joining the points must be equal to 0 sq.units.

[Why so? Because if you join any three random points lying on a line, and calculate their area, it's equal to 0 since they lie on the exact same line.]

Now, let's use the area of a triangle formula for any three coordinates to find out the area of the three points to determine it's collinearity.

$\sf Area \ of \ a \ \Delta gle = \dfrac{1}{2} \Bigg\{ x_1\left(y_2 - y_3\right) + x_2\left(y_3 - y_1\right) + x_3\left(y_1 - y_2\right) \Bigg\}$

Co-ordinates are:

x₁ = 0

x₂ = 2

x₃ = 3

y₁ = 3

y₂ = 1

y₃ = -1

$\dashrightarrow \ \sf Area \ of \ a \ \Delta gle = \dfrac{1}{2} \Bigg\{x_1\left(y_2 - y_3\right) + x_2\left(y_3 - y_1\right) + x_3\left(y_1 - y_2\right) \Bigg\}$

$\dashrightarrow \ \sf Area \ of \ a \ \Delta gle = \dfrac{1}{2} \Bigg\{0\left(1 - (-1)\right) + 2\left(-1 - 3\right) + 3\left(3 - 1\right) \Bigg\}$

$\dashrightarrow \ \sf Area \ of \ a \ \Delta gle = \dfrac{1}{2} \Bigg\{0 + 2\left(-4\right) + 3\left(2\right) \Bigg\}$

$\dashrightarrow \ \sf Area \ of \ a \ \Delta gle = \dfrac{1}{2} \Bigg\{-8 + 6 \Bigg\}$

$\dashrightarrow \ \sf Area \ of \ a \ \Delta gle = \dfrac{1}{2} \Bigg\{-2 \Bigg\}$

$\dashrightarrow \ \sf Area \ of \ a \ \Delta gle = -1$

Area can't be negative, therefore;

$\dashrightarrow \ \underline{\underline{\sf{Area \ of \ a \ \Delta gle = 1 \ sq.units}}}$

The area isn't equal to 0 sq.units, therefore the given co-ordinates are not collinear.

Answer 2

Answer:

Collinear points are those points that lie on the same line.

For three or more co-ordinates to be collinear, the area formed by joining the points must be equal to 0 sq.units.

[Why so? Because if you join any three random points lying on a line, and calculate their area, it's equal to 0 since they lie on the exact same line.]

Now, let's use the area of a triangle formula for any three coordinates to find out the area of the three points to determine it's collinearity.

$\sf Area \ of \ a \ \Delta gle = \dfrac{1}{2} \Bigg\{ x_1\left(y_2 - y_3\right) + x_2\left(y_3 - y_1\right) + x_3\left(y_1 - y_2\right) \Bigg\}$

Co-ordinates are:

x₁ = 0

x₂ = 2

x₃ = 3

y₁ = 3

y₂ = 1

y₃ = -1

$\dashrightarrow \ \sf Area \ of \ a \ \Delta gle = \dfrac{1}{2} \Bigg\{x_1\left(y_2 - y_3\right) + x_2\left(y_3 - y_1\right) + x_3\left(y_1 - y_2\right) \Bigg\}$

$\dashrightarrow \ \sf Area \ of \ a \ \Delta gle = \dfrac{1}{2} \Bigg\{0\left(1 - (-1)\right) + 2\left(-1 - 3\right) + 3\left(3 - 1\right) \Bigg\}$

$\dashrightarrow \ \sf Area \ of \ a \ \Delta gle = \dfrac{1}{2} \Bigg\{0 + 2\left(-4\right) + 3\left(2\right) \Bigg\}$

$\dashrightarrow \ \sf Area \ of \ a \ \Delta gle = \dfrac{1}{2} \Bigg\{-8 + 6 \Bigg\}$

$\dashrightarrow \ \sf Area \ of \ a \ \Delta gle = \dfrac{1}{2} \Bigg\{-2 \Bigg\}$

$\dashrightarrow \ \sf Area \ of \ a \ \Delta gle = -1$

Area can't be negative, therefore;

$\dashrightarrow \ \underline{\underline{\sf{Area \ of \ a \ \Delta gle = 1 \ sq.units}}}$

The area isn't equal to 0 sq.units, therefore the given co-ordinates are not collinear.