Question

show that point A(-6,4)B(-13,4)C(-6,-3)D(13,-3)

Answer 1

Show that point A(-6,4)B(-13,4)C(-6,-3)D(-13,-3) are the vertices of a square

The equation of the line passing through the points A and B is $y = 4$ , so this line is parallel to the x-axis.

The equation of the line passing through the points C and D is $y = -3$ , so this line is parallel to the x-axis.

The equation of the line passing through the points B and D is $x = -13$ , so this line is parallel to the x-axis.

The equation of the line passing through the points A and C is $y = -6$ , so this line is parallel to the x-axis.

So

1) the line AB is parallel to the line CD and the line BD is parallel to the line AC, hence ABDC is a parallelogram,

2) the line AB is perpendicular to the line BD, hence ABDC is a rectangle,

3) $AB = |-13 - (-6)| = 7$

$BD = |-3 - 4| = 7$

$CD = |-13 - (-6)| = 7$

$AC = |-3 - 4| = 7$ , so $\mathrm{AB} = \mathrm{BD} = \mathrm{CD} = \mathrm{AC}$ , hence ABDC is a rhombus.

ABCD is a square since it is a rhombus and a rectangle