The coordinates of the vertices of △ABC are A(1,4), B(-2,-1), and C(-3,-2). The height h to side BC is √2, and b is the length of BC. What is the area of △ABC?
I don't really know how to approach or start this problem, so if you can help, please do! Thanks!
I'm not really sure if I should use the distance formula to find the distance between the 3 points or do something else. Thanks!
T
To find the area of △ABC, we can use the formula:
Area = 1/2 * base * height
First, let's find the length of the base, BC. We can use the distance formula:
Distance = √((x2 - x1)^2 + (y2 - y1)^2)
Let's calculate the distance between points B(-2,-1) and C(-3,-2):
Distance BC = √((-3 - (-2))^2 + (-2 - (-1))^2)
= √((-3 + 2)^2 + (-2 + 1)^2)
= √((-1)^2 + (-1)^2)
= √(1 + 1)
= √2
Now we have the length of the base, BC, which is √2.
Next, let's find the area using the formula:
Area = 1/2 * BC * h
Substituting the values we have:
Area = 1/2 * √2 * √2
= 1/2 * 2
= 1
Therefore, the area of △ABC is 1 square unit.
BC is the base , and h is the height
area of a triangle is ... 1/2 * b * h ... 1/2 * BC * √2
use the distance formula to find BC