Chapter 8.3, Problem 35WE

To determine

To Explain: whether the points $\left(0,-1\right),\left(1,-2\right),\left(2,-4\right),\left(-1,0\right)$ are collinear or not.

Answer to Problem 35WE

The points are not collinear.

Explanation of Solution

Given information:

The points of a plane are given as:

  $\left(0,-1\right),\left(1,-2\right),\left(2,-4\right),\left(-1,0\right)$

Formula used:

To check the collinear points,

  $\frac{x_1-x_2}{x_1-x_3}=\frac{y_1-y_2}{y_1-y_3}=\frac{z_1-z_2}{z_1-z_3}$

The three or more points are collinear when they are in straight line.

We find here with the help of parallelogram formula:

  $AB=\sqrt{{\left(x_1-x_2\right)}^2+{\left(y_1-y_2\right)}^2}$

Such as $AB=BC=CD=DA$

Consider the points,

  $\left(0,-1\right),\left(1,-2\right),\left(2,-4\right),\left(-1,0\right)$

Recall that to check the collinear points,

We find here with the help of parallelogram formula:

  $AB=\sqrt{{\left(x_1-x_2\right)}^2+{\left(y_1-y_2\right)}^2}$

Such as $AB=BC=CD=DA$

Therefore, the points are given by,

  $\left(0,-1\right),\left(1,-2\right),\left(2,-4\right),\left(-1,0\right)$

Make a table of the coordinates in increasing order $x$ -coordinates then change in the $x$ -coordinates and $y$ -coordinates as move one point to the next.

If the ratio of vertical change divide by horizontal change is constant, the points are collinear. Otherwise they are not.

Check the points are collinear or not.

  $\begin{array}{cc}x& y\\ 0& -1\\ 1& -2\\ 2& -4\\ -1& 0\end{array}$

Here, first we check with the help of formula:

As the values of:

  $\begin{array}{l}\left(x_1,y_1\right)=\left(0,-1\right)\\ \left(x_2,y_2\right)=\left(1,-2\right)\\ \left(x_3,y_3\right)=\left(2,-4\right)\\ \left(x_4,y_4\right)=\left(-1,0\right)\end{array}$

Now apply these values in the formula of parallelogram:

We find here with the help of parallelogram formula:

  $AB=\sqrt{{\left(x_1-x_2\right)}^2+{\left(y_1-y_2\right)}^2}$

Such as $AB=BC=CD=DA$

  $\begin{array}{l}AB=\sqrt{{\left(x_1-x_2\right)}^2+{\left(y_1-y_2\right)}^2}\\ AB=\sqrt{{\left(0-1\right)}^2+{\left(-1-2\right)}^2}\\ AB=\sqrt{-1^2+{\left(-3\right)}^2}\\ AB=\sqrt{1+9}\\ AB=\sqrt{10}\end{array}$

Now find the line $BC$ .

  $\begin{array}{l}BC=\sqrt{{\left(x_2-x_3\right)}^2+{\left(y_2-y_3\right)}^2}\\ BC=\sqrt{{\left(1-2\right)}^2+{\left(-2+4\right)}^2}\\ BC=\sqrt{-1^2+{\left(2\right)}^2}\\ BC=\sqrt{1+4}\\ BC=\sqrt{5}\end{array}$

Now find the line $CD$ .

  $\begin{array}{l}CD=\sqrt{{\left(x_3-x_4\right)}^2+{\left(y_3-y_4\right)}^2}\\ CD=\sqrt{{\left(2+1\right)}^2+{\left(-4-0\right)}^2}\\ CD=\sqrt{3^2+{\left(-4\right)}^2}\\ CD=\sqrt{9+16}\end{array}$

The value of $CD$ is:

  $\begin{array}{l}CD=\sqrt{25}\\ CD=5\end{array}$

Now find the line $DA$ .

  $\begin{array}{l}DA=\sqrt{{\left(x_4-x_1\right)}^2+{\left(y_4-y_1\right)}^2}\\ DA=\sqrt{{\left(-1-0\right)}^2+{\left(0+1\right)}^2}\\ DA=\sqrt{-1^2+{\left(1\right)}^2}\\ DA=\sqrt{1+1}\end{array}$

The value of $DA$ is:

  $\begin{array}{l}DA=\sqrt{1+1}\\ DA=\sqrt{2}\end{array}$

Here, $AB\ne BC\ne CD\ne DA$

Because all the points are not lie in same line, so it is not collinear.

Thus, the points are not collinear.