Question

determine whether the points x, y, and z can be the vertices of a triangle. if so, classify the triangle as acute, right, or obtuse.

1. x(0,0), y(4,7), z(8,2)
2. x(-3,-1), y(1,4), z(5,-2)
3. x(3,8), y(9,6), z(6,2)
4. x(-4,2), y(-1,-3), z(3,1)

Explanation:

For Problem 1: X(0,0), Y(4,7), Z(8,2)

Step1: Calculate side lengths squared

Use distance formula squared: $d^2=(x_2-x_1)^2+(y_2-y_1)^2$
$XY^2=(4-0)^2+(7-0)^2=16+49=65$
$YZ^2=(8-4)^2+(2-7)^2=16+25=41$
$XZ^2=(8-0)^2+(2-0)^2=64+4=68$

Step2: Check triangle inequality

Verify sum of any two sides > third:
$\sqrt{65}+\sqrt{41}>\sqrt{68}$, $\sqrt{65}+\sqrt{68}>\sqrt{41}$, $\sqrt{41}+\sqrt{68}>\sqrt{65}$ (all true)

Step3: Classify triangle via Pythagoras

Compare $XZ^2$ to $XY^2+YZ^2$:
$68 < 65+41=106$

Step1: Calculate side lengths squared

$XY^2=(1-(-3))^2+(4-(-1))^2=16+25=41$
$YZ^2=(5-1)^2+(-2-4)^2=16+36=52$
$XZ^2=(5-(-3))^2+(-2-(-1))^2=64+1=65$

Step2: Check triangle inequality

$\sqrt{41}+\sqrt{52}>\sqrt{65}$, $\sqrt{41}+\sqrt{65}>\sqrt{52}$, $\sqrt{52}+\sqrt{65}>\sqrt{41}$ (all true)

Step3: Classify triangle via Pythagoras

$65 < 41+52=93$

Step1: Calculate side lengths squared

$XY^2=(9-3)^2+(8-8)^2=36+0=36$
$YZ^2=(6-9)^2+(2-8)^2=9+36=45$
$XZ^2=(6-3)^2+(2-8)^2=9+36=45$

Step2: Check triangle inequality

$\sqrt{36}+\sqrt{45}>\sqrt{45}$, $\sqrt{45}+\sqrt{45}>\sqrt{36}$ (all true)

Step3: Classify triangle via Pythagoras

$45 < 36+45=81$; two sides equal

Answer:

This is an acute triangle.

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For Problem 2: X(-3,-1), Y(1,4), Z(5,-2)