Chapter 1.8, Problem 41E

To determine

To verify: The given points are the vertices of a square.

Explanation of Solution

Given information:

The four points $\text{A}\left(-2,9\right),\text{B}\left(4,6\right),\text{C}\left(1,0\right)\text{ and D}\left(-5,3\right)$ .

Formula used:

Distance formula between two points $\text{X}\left(a_1,b_1\right)$ and $\text{Y}\left(a_2,b_2\right)$ is calculated as,

  $d\left(\text{XY}\right)=\sqrt{{\left(a_2-a_1\right)}^2+{\left(b_2-b_1\right)}^2}$

A rhombus is a special kind of quadrilateral with all sides equal and square is a rhombus with equal lengths of diagonals.

Proof:

Consider the given four points $\text{A}\left(-2,9\right),\text{B}\left(4,6\right),\text{C}\left(1,0\right)\text{ and D}\left(-5,3\right)$ .

Plot the given points on coordinate plane and join the vertices to form the quadrilateral ABCD as,

  

From the above figure, calculate the distance between all the four points A, B, C and D using distance formula to measure the length of sides.

Recall that the distance formula between two points $\text{X}\left(a_1,b_1\right)$ and $\text{Y}\left(a_2,b_2\right)$ is calculated as,

  $d\left(\text{XY}\right)=\sqrt{{\left(a_2-a_1\right)}^2+{\left(b_2-b_1\right)}^2}$

So, lengths of sides AB, BC, CA and DA will be calculated as,

  $\begin{array}{c}d\left(\text{AB}\right)=\sqrt{{\left(4-\left(-2\right)\right)}^2+{\left(6-9\right)}^2}\\ =\sqrt{{\left(4+2\right)}^2+{\left(-3\right)}^2}\\ =\sqrt{36+9}\\ =\sqrt{45}\end{array}$

  $\begin{array}{c}d\left(\text{BC}\right)=\sqrt{{\left(1-4\right)}^2+{\left(0-6\right)}^2}\\ =\sqrt{{\left(-3\right)}^2+{\left(-6\right)}^2}\\ =\sqrt{9+36}\\ =\sqrt{45}\end{array}$

  $\begin{array}{c}d\left(\text{CD}\right)=\sqrt{{\left(-5-1\right)}^2+{\left(3-0\right)}^2}\\ =\sqrt{{\left(-6\right)}^2+{\left(3\right)}^2}\\ =\sqrt{36+9}\\ =\sqrt{45}\end{array}$

  $\begin{array}{c}d\left(\text{DA}\right)=\sqrt{{\left(-5-\left(-2\right)\right)}^2+{\left(9-3\right)}^2}\\ =\sqrt{{\left(-5+2\right)}^2+{\left(6\right)}^2}\\ =\sqrt{9+36}\\ =\sqrt{45}\end{array}$

Recall that a rhombus is a special kind of quadrilateral with all sides equal and square is a rhombus with equal lengths of diagonals.

Since, all the sides are equal, so, the given points are of rhombus.

Now, calculate the lengths of diagonals AC and BD,

  $\begin{array}{c}d\left(\text{AC}\right)=\sqrt{{\left(1-\left(-2\right)\right)}^2+{\left(0-9\right)}^2}\\ =\sqrt{{\left(1+2\right)}^2+{\left(9\right)}^2}\\ =\sqrt{9+81}\\ =\sqrt{90}\end{array}$

  $\begin{array}{c}d\left(\text{BD}\right)=\sqrt{{\left(-5-4\right)}^2+{\left(3-6\right)}^2}\\ =\sqrt{{\left(-9\right)}^2+{\left(-3\right)}^2}\\ =\sqrt{81+9}\\ =\sqrt{90}\end{array}$

Since, the diagonals are also of equal lengths, so, this rhombus is a square.

Since, the quadrilateral formed from the given points has equal sides and equal diagonals, thus it is proved that the given points are the vertices of a square.