Answer:
Explanation:
(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.
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(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