Question

Do the points (3 , 2) , (-2, -3) and (2 ,3) form a triangle, if so name the triangle formed ?

Answer 1

Given :

We are given three points (3,2), (-2,-3) and (2,3).

To find :

We've to find whether these points form a triangle or not. If yes, we should also ne the triangle.

Solution :

Let the points be :

• K = (3,2)
• L = (-2,-3)
• M = (2,3)

Let the sides be :

• KL
• LM
• MK

Here, we'll use the distance formula to find distances between the given points.

$\bigstar \: \: \boxed{ \sqrt{ {(x_2-x_1)}^{2} + {(y_2-y_1)}^{2} } } \: \: \bigstar$

• Finding KL :

$\implies KL = \: \sqrt{(3 + 2)^2+( - 3 - 2)^2}$

$\implies KL= \sqrt{50} \: units$

• Finding LM :

$\implies LM = \sqrt{ {( 2 + 2) }^{2} + {(3 + 3)}^{2} }$

$\implies LM = \sqrt{52} \: units$

• Finding MK :

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

$\implies MK= \sqrt{2} \: units$

Here, we see that the sum of any two sides(distances) is greater than the third distance, so it is a triangle.

• KL + LM > MK
• LM + MK > KL
• MK + KL > LM

Also, here, we see that KL²+MK²=LM² by pythagoras theorem :

$\implies { \sqrt{50} }^{2} + { \sqrt{2} }^{2} = { \sqrt{52} }^{2}$

$\implies 50+2=52$

$\implies 52=52$

LHS = RHS

_________________________

Therefore, ∆KLM is a right-angled triangle.

@VenusVeronica

Answer 2

Let us apply the distance formula to find the distances PQ, QR and PR, where P(3, 2), Q(–2, –3) and R(2, 3) are the given points. We have Since the sum of any two of these distances is greater than the third distance, therefore, the points P, Q and R form a triangle. Also, PQ2 + PR2 = QR2, by the converse of Pythagoras theorem, we have ∠P = 90°. Therefore, PQR is a right triangle.