Question

Triangle XYZ is shown on the coordinate grid.

Answer 1

Solution:

Distance between two points $(x_{1},y_{1}), (x_{2},y_{2})$ in two dimensional coordinate plane is given by: $\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2$

$XY=\sqrt{(4+1)^2+(4-5)^2}\\\\ XY=\sqrt{25 +1}\\\\ XY=\sqrt{26}\\\\ YZ=\sqrt{(4+2)^2+(4-0)^2}\\\\ YZ=\sqrt{36 +16}\\\\ YZ=\sqrt{52}\\\\ YZ=2\sqrt{13}\\\\ XZ=\sqrt{(-2+1)^2+(5-0)^2}\\\\ XZ=\sqrt{1+25}\\\\ XZ=\sqrt{26}\\\\ Also, (XY)^2+(XZ)^2=(\sqrt{26})^2+(\sqrt{26})^2=52=(YZ)^2$

As, Converse of Pythagorean theorem states that if sum of squares of any two adjacent sides in a triangle is equal to Square of length of third side, then the triangle is right angled triangle.

The true statement about the triangle XYZ is

1. XY measures √26units.

2. XYZ is an isosceles triangle.

3. XYZ is a right triangle.

Answer 2

Answer:

XY measures √26units.

XYZ is an isosceles triangle.

XYZ is a right triangle.

Step-by-step explanation:

The given triangle has vertices $X(-1,5),Y(4,4),(-2,0)$

We now apply the distance formula to find the length of all sides to obtain;

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

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

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

$|XY|=\sqrt{25+1}$

$|XY|=\sqrt{26}$

The length of YZ is

$|YZ|=\sqrt{(4--2)^2+(4-0)^2}$

$|YZ|=\sqrt{(4+2)^2+(4-0)^2}$

$|YZ|=\sqrt{6^2+(4)^2}$

$|YZ|=\sqrt{36+16}$

$|YZ|=\sqrt{52}$

The length of ZX is

$|ZX|=\sqrt{(-2--1)^2+(0-5)^2}$

$|ZX|=\sqrt{(-2+1)^2+(0-5)^2}$

$|ZX|=\sqrt{(-1)^2+(-5)^2}$

$|ZX|=\sqrt{1+25}$

$|ZX|=\sqrt{26}$

Since

$|XY|=\sqrt{26}=|ZX|$

The triangle is isosceles.

Slope of XZ

$\frac{5-0}{-1--2}=\frac{5}{1} =5$

Slope of XY

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

Since

$\frac{-1}{5}\times 5=-1$

The triangle is a right angle triangle.