Question

1. Find the length of the sides of the triangle whose vertices are A (3,4) B(2,-1) C (4,-6)
2.prove that the points (2,-2), (-2,1) and (5,2) are the vertices of a right angled triangle

do in your copy then post it plzzz fast

Answer 1

hey mate
here's the solution

Answer 2

(1)

Given vertices are A(3,4), B(2,-1) and C(4,-6).

We need to calculate the length of the sides AB,BC,AC.

We have to distance formula which can tell the distance between 2 points.

d = √(x2 - x1)^2 + (y2 - y1)^2.

(1)

AB = √(2 - 3)^2 + (-1 - 4)^2

     = √(-1)^2 + (5)^2

     = √1 + 25

     = √26.

(2)

BC = √(4 - 2)^2 + (-6 + 1)^2

     = √(2)^2 + (5)^2

     = √4 + 25

     = √29

(3)

AC = √(4 - 3)^2 + (-6 - 4)^2

     = √(1)^2 + (-10)^2

     = √1 + 100

     = √101

Therefore, the length of the sides of the triangle are √26,√29 and √101.

----------------------------------------------------------------------------------------------------------

(2)

Let the given points be A(2,-2), B(-2,1) and C(5,2).

Using the distance formula,w e find that

⇒ AB = √(-2 - 2)^2 + (1 + 2)^2

         = √16 + 9

         = √25.

⇒ BC = √(5 + 2)^2 + (2 - 1)^2

          = √49 + 1

          = √50.

⇒ AC = √(5 - 2)^2 + (2 + 2)^2

          = √9 + 16

          = √25.

       

Now,

⇒ AB^2 + AC^2

⇒ (5)^2 + (5)^2

⇒ 25 + 25

⇒ 50.

⇒ (BC)^2.

Therefore, AB^2 + AC^2 = BC^2.

∴ We can conclude that ΔABC is a right angled triangle.

Hope it helps!