Question

prove that the points A(-3,0), B(1,-3 ) and (4,1) are the vertices of an isosceles right angle triangle. find the area of the triangle.

Answer 1

Sol:
Vertices of the triangle are A(-3, 0), B(1, -3), C(4, 1)
Distance between two points =  (x2 - x1)2+(y2 - y1)2 
AB = √[(1 + 3)2 + (-3 -0)2] = 5
BC = √[(4 - 1)2 + (1 + 3)2] = 5
AC = √[(4 + 3)2 + (1 -0)2] = 50 = 52 
AB = BC
Therefore, ΔABC is an isosceles triangle.
52 + 52 = (52 )2
⇒ 50 = 50
⇒ (AB)2 + (BC)2 = (AC)2
So, the triangle satisfies the pythagoras theorem and hence it is a right angled triangle.

Area of the isosceles right angled triangle = 1 2 ×base × height = 1 2 × 5 × 5 = 12.5 sq units.

Answer 2

Answer:

It is a right angled triangle & the area is 12.5 units²

Step-by-step explanation:

Vertices of the triangle are A(-3, 0), B(1, -3), C(4, 1)

Distance between two points =  (x2 - x1)2+(y2 - y1)2 

AB = √[(1 + 3)2 + (-3 -0)2] = 5

BC = √[(4 - 1)2 + (1 + 3)2] = 5

AC = √[(4 + 3)2 + (1 -0)2] = 50 = 52 

AB = BC

Therefore, ΔABC is an isosceles triangle.

52 + 52 = (52 )2

⇒ 50 = 50

⇒ (AB)2 + (BC)2 = (AC)2

So, the triangle satisfies the pythagoras theorem and hence it is a right angled triangle.

Area of the isosceles right angled triangle = 1 2 ×base × height = 1 2 × 5 × 5 = 12.5 sq units.

HOPE THIS HELP YOU!!!

PLEASE MARK ME AS THE BRAINLIEST