Question

Determine and prove what shape is formed for the given coordinates for ABCD, and then find the

Answer 1

Answer:

Square

Step-by-step explanation:

Plot points A (10,-4), B (6,-7), C (3,-3), D (7,0) on the coordinate plane.

Find slopes of the sides:

$AB:\ \dfrac{-7-(-4)}{6-10}=\dfrac{3}{4}\\ \\BC:\ \dfrac{-3-(-7)}{3-6}=-\dfrac{4}{3}\\ \\CD:\ \dfrac{0-(-3)}{7-3}=\dfrac{3}{4}\\ \\DA:\ \dfrac{-4-0}{10-7}=-\dfrac{4}{3}$

The slopes of opposite sides are the same, so opposite sides are parallel. The slopes of adjacent sides have product of -1, then adjacent sides are perpendicular.

Find the lengths of all sides:

$AB=\sqrt{(10-6)^2+(-4-(-7))^2}=\sqrt{4^2+3^2}=\sqrt{16+9}=\sqrt{25}=5\ units\\ \\BC=\sqrt{(6-3)^2+(-7-(-3))^2}=\sqrt{3^2+4^2}=\sqrt{16+9}=\sqrt{25}=5\ units\\ \\CD=\sqrt{(3-7)^2+(-3-0)^2}=\sqrt{4^2+3^2}=\sqrt{16+9}=\sqrt{25}=5\ units\\ \\AD=\sqrt{(10-7)^2+(-4-0)^2}=\sqrt{3^2+4^2}=\sqrt{16+9}=\sqrt{25}=5\ units$

All four sides are of the same length.

Quadrilateral ABCD is a square (all sides of equal length and perpendicular)