find the area of a right angled riangle. find the distance from A to B. Find the distance from B to C. Find the midpoint
B(-1,4)
A(-1,-1)
C(4,-1)
thanks for your time and help.
AB = 5
BC = √(-5)^2 + )5^2) = √50 = 5√2
what midpoint?
how did you find the answer? did you take the slope??? can you explain how
To find the area of a right-angled triangle, you can use the formula: Area = (base * height) / 2.
First, let's find the base and height of the triangle. The base is the distance from point A to point C, and the height is the distance from point A to point B. We'll use the distance formula to find these distances.
Distance Formula: The distance between two points (x₁, y₁) and (x₂, y₂) is given by the formula: distance = √((x₂ - x₁)² + (y₂ - y₁)²).
Let's calculate the distances:
1. Distance from A to C:
A(-1, -1)
C(4, -1)
Using the distance formula:
distance_AC = √((4 - (-1))² + (-1 - (-1))²)
= √(5² + 0²)
= √(25 + 0)
= √25
= 5
The base of the triangle, AC, is 5 units.
2. Distance from A to B:
A(-1, -1)
B(-1, 4)
Using the distance formula:
distance_AB = √((-1 - (-1))² + (4 - (-1))²)
= √(0² + 5²)
= √(0 + 25)
= √25
= 5
The height of the triangle, AB, is also 5 units.
Now let's calculate the area of the right-angled triangle using the formula mentioned earlier:
Area = (base * height) / 2
= (5 * 5) / 2
= 25 / 2
= 12.5 square units
Therefore, the area of the right-angled triangle is 12.5 square units.
To find the distance from point A to point B, we have already found that it is 5 units.
To find the distance from point B to point C, we can use the distance formula again:
Using the distance formula:
distance_BC = √((4 - (-1))² + (-1 - 4)²)
= √(5² + (-5)²)
= √(25 + 25)
= √50
= 5√2
Therefore, the distance from point B to point C is 5√2 units.
To find the midpoint of line segment BC, we can use the midpoint formula:
Midpoint Formula: The coordinates of the midpoint between two points (x₁, y₁) and (x₂, y₂) are given by the formula: ( (x₁ + x₂) / 2 , (y₁ + y₂) / 2 ).
Let's find the midpoint:
B(-1, 4)
C(4, -1)
Using the midpoint formula:
midpoint_BC = ( ( -1 + 4 ) / 2 , ( 4 + (-1) ) / 2 )
= ( 3 / 2 , 3 / 2 )
Therefore, the midpoint of line segment BC is (3/2, 3/2).
I hope this explanation was clear and helpful!