Place the square in the corner of the first quadrant of the coordinate plane.
See solution.
Practice makes perfect
We are asked to show that the diagonals of a square are perpendicular.
This statement is true because a square is a rhombus, but let's see how we can prove this using coordinates as asked.
Let's place the square in the corner, at the origin, of the first quadrant of the coordinate plane so that it's sides are parallel to the coordinate axes.
Since the sides are congruent and the angles are right angles, if vertex D has coordinates (d,0), then vertex B has coordinates (0,d) and vertex C has coordinates (d,d).
The diagonals are perpendicular if we can show that their slopes multiply to - 1.
Let's use the Slope Formula to find the slope of the diagonals.
2
&Slope ofAC: &&m_(AC)=d-0/d-0=1
&Slope ofBD: &&m_(BD)=0-d/d-0=- 1
Since m_(AC) * m_(BD)=1(- 1)=-1, the slopes multiply to - 1, so the diagonals are perpendicular.
Extra
Applying Real Numbers to the Same Problem
For the sake of understanding. Let's replace d with 3, and then see what we get.
Then, let's use the Slope Formula to show that the slope of the diaganols when multiplied to each other will equal - 1. 2
&Slope ofAC: &&m_(AC)=3-0/3-0=1
&Slope ofBD: &&m_(BD)=0-3/3-0=- 1
Since m_(AC) * m_(BD)=1(- 1)=-1, the slopes multiply to - 1, so the diagonals are perpendicular.
