# Thread: Analytic Geometry Q2

1. ## Analytic Geometry Q2

Question:
Show that $A(-2,1)$ , $B(5,-2)$ and $C(3,3)$ are vertices of an equilateral triangle.

Attempt:

Distance between points $A(-2,1)$ and $B(5,-2)$:

$AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
$AB = \sqrt{(5-(-2))^2 + (-2-1)^2}$
$AB = \sqrt{49 + 9}$
$AB = \sqrt{58}$

Distance between points $B(5,-2)$ and $C(3,3)$:

$BC = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
$BC = \sqrt{(3-5)^2+(3-(-2))^2}$
$BC = \sqrt{4 + 25}$
$BC = \sqrt{29}$

Distance between points $A(-2,1)$ and $C(3,3)$:

$AC = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
$AC = \sqrt{(3-(-2))^2 + (3-1)^2}$
$AC = \sqrt{25 + 4}$
$AC = \sqrt{29}$

I have found out the distance between the points. How can I find out if it is a right angled triangle?

2. This is not an equilateral triangle( all the sides are not equal)
But it's a right triangle because
AB^2= BC^2+ AC^2 (Pythagoras theorem)

Pythatgoras Theorem states that the square of the hypotnuse(side opp. to right angle) of a right triangle is equal to the sum of squares of sides

3. Originally Posted by looi76
How can I find out if it is a right angled triangle?
Well, most easily by using Pythagorean theorem, so it is because $\overline{AC}^2+\overline{BC}^2=\overline{AB}^2\Le ftrightarrow 29+29=58$.