Question

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

Answer 1

Answer:

Is an acute triangle

Step-by-step explanation:

we have

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

so

The polygon is a triangle

we know that

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

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

Remember 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

$c^{2}$ -----> is an acute triangle

where

c is the greater side

step 1

Find the distance GH

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

substitute

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

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

$GH=\sqrt{13}\ units$

step 2

Find the distance HI

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

substitute

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

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

$HI=\sqrt{17}\ units$

step 3

Find the distance GI

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

substitute

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

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

$GI=\sqrt{20}\ units$

step 4

Let

$c=GI=\sqrt{20}\ units$

$a=HI=\sqrt{17}\ units$

$b=GH=\sqrt{13}\ units$

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

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

Compare

$20 < 30$

therefore

Is an acute triangle