The coordinates of (triangle) ABC are A(-3,-1 ) , B( 5,1 ) and C (2,4)
(i) Calculate the length of AB , BC and AC
(ii) Hence, show that (triangle) is a right- angled triangle
AB^2 = 64 + 4 = 68
BC^2 = 9 + 9 = 18
AC^2 = 25 + 25 = 50
Note that 68 = 50 + 18
therefore a right triangle
To calculate the length of each side of the triangle, we can use the distance formula:
Distance between two points (x1, y1) and (x2, y2) = √((x2 - x1)^2 + (y2 - y1)^2)
(i) Let's calculate the length of each side:
AB:
A(-3,-1) and B(5,1)
Distance AB = √((5 - (-3))^2 + (1 - (-1))^2)
= √((5 + 3)^2 + (1 + 1)^2)
= √(8^2 + 2^2)
= √(64 + 4)
= √68
≈ 8.246
BC:
B(5,1) and C(2,4)
Distance BC = √((2 - 5)^2 + (4 - 1)^2)
= √((-3)^2 + (3)^2)
= √(9 + 9)
= √18
≈ 4.242
AC:
A(-3,-1) and C(2,4)
Distance AC = √((2 - (-3))^2 + (4 - (-1))^2)
= √((2 + 3)^2 + (4 + 1)^2)
= √(5^2 + 5^2)
= √(25 + 25)
= √50
≈ 7.071
(ii) Now, let's check if the triangle is right-angled.
We can determine if a triangle is right-angled by checking if the square of the longest side is equal to the sum of the squares of the other two sides.
In this case, the longest side is AC (approximately 7.071), and the squares of the other two sides are:
AB^2 = (approximately 8.246)^2 ≈ 68
BC^2 = (approximately 4.242)^2 ≈ 18
Checking if AC^2 = AB^2 + BC^2:
7.071^2 ≈ 50 ≈ 68 + 18
Since 50 is equal to 68 + 18, we can conclude that the triangle ABC is a right-angled triangle.