Question

Lesson 4.10 - Analytic Geometry

Answer 1

Geometric shapes are mathematical shapes that include squares and triangles

How to prove the shape is a square

The side lengths of a square are congruent, and the adjacent sides are perpendicular.

So, we start by calculating the side lengths using the following distance formula

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

Using the above formula, we have:

$AB = \sqrt{(3-2)^2 + (4+2)^2} = \sqrt{37$

$BC = \sqrt{(2+4)^2 + (-2+1)^2} = \sqrt{37$

$CD = \sqrt{(-4+3)^2 + (-1-5)^2} = \sqrt{37}$

$DA = \sqrt{(-3-3)^2 + (5-4)^2} = \sqrt{37}$

The above shows that the side lengths of the square are congruent.

Next, calculate the slope of the sides using:

$m = \frac{y_2 -y_1}{x_2 -x_1}$

So, we have:

$m_{AB} = \frac{4 +2}{3 -2} = 6$

$m_{BC} = \frac{-2 + 1}{2+4} = -\frac 16$

$m_{CD} = \frac{-1 -5}{-4+3} = 6$

$m_{DA} = \frac{5 -4}{-3-3} = -\frac 16$

Notice that the opposite slopes are congruent, and the adjacent slopes are opposite reciprocal.

The above highlight, and the equal side lengths show that the figure (1) is a square

How to prove the shape is a right isosceles triangle

The legs of a right isosceles triangle are congruent, and the legs are perpendicular.

So, we start by calculating the lengths of the legs using the following distance formula

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

Using the above formula, we have:

$XY = \sqrt{(5-4)^2 + (-1-4)^2} = \sqrt{26$

$XZ = \sqrt{(5-0)^2 + (-1+2)^2} = \sqrt{26$

The above shows that the legs of the right isosceles triangle are congruent.

Next, calculate the slope of the legs using:

$m = \frac{y_2 -y_1}{x_2 -x_1}$

So, we have:

$m_{XY} = \frac{5-4}{-1 -4} = -\frac{1}{5}$

$m_{XZ} = \frac{5-0}{-1+2} = 5$

Notice that the slopes are opposite reciprocal.

The above highlight, and the equal legs show that the figure (2) is a right isosceles triangle

Read more about geometric shapes at:

brainly.com/question/14285697