Question

The coordinates G(7,3), H(9, 0), (5, -1) form what type of polygon?

Answer 1

Answer:

ACUTE !!!!!!!!

Step-by-step explanation:

Answer 2

Answer:

Is an acute triangle

Step-by-step explanation:

we know that

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

we have

G(7,3), H(9, 0), I(5, -1)

step 1

Find the distance GH

substitute in the formula

$d=\sqrt{(0-3)^{2}+(9-7)^{2}}$

$d=\sqrt{(-3)^{2}+(2)^{2}}$

$GH=\sqrt{13}\ units$

step 2

Find the distance IH

substitute in the formula

$d=\sqrt{(0+1)^{2}+(9-5)^{2}}$

$d=\sqrt{(1)^{2}+(4)^{2}}$

$IH=\sqrt{17}\ units$

step 3

Find the distance GI

substitute in the formula

$d=\sqrt{(-1-3)^{2}+(5-7)^{2}}$

$d=\sqrt{(-4)^{2}+(-2)^{2}}$

$GI=\sqrt{20}\ units$

step 4

Verify what type of triangle is the polygon

we know that

If applying the Pythagoras Theorem

$c^{2}=a^{2}+b^{2}$ ----> is a right triangle

$c^{2}> a^{2}+b^{2}$ ----> is an obtuse triangle

$c^{2}< a^{2}+b^{2}$ ----> is an acute triangle

where

c is the greater side

we have

$c=\sqrt{20}\ units$

$a=\sqrt{17}\ units$

$b=\sqrt{13}\ units$

substitute

$c^{2}= (\sqrt{20})^{2}=20$

$a^{2}+b^{2}=(\sqrt{17})^{2}+(\sqrt{13})^{2}=30$

therefore

$c^{2}< a^{2}+b^{2}$

Is an acute triangle