Find the area of a triangle ABC with vertices

Question

Find the area of a triangle ABC with vertices

A(2, 3), B(4, 7) and C(7, 3)
A. 20 1
2
𝑢𝑛𝑖𝑡𝑠2.
B. √10 𝑢𝑛𝑖𝑡𝑠2
C. 10 1
2
𝑢𝑛𝑖𝑡𝑠2
D. 10 𝑢𝑛𝑖𝑡𝑠2

Answer 1

We can use the formula for the area of a triangle given its vertices:

Answer 2

To find the area of a triangle with given vertices, you can use the formula for the area of a triangle:

Area = (1/2) * base * height

where the base is the length of any one side of the triangle and the height is the perpendicular distance from the base to the opposite vertex.

In this case, let's consider AB as the base and the point C as the opposite vertex. To find the length of AB, we can use the distance formula:

Length of AB = √((x2 - x1)^2 + (y2 - y1)^2) = √((4 - 2)^2 + (7 - 3)^2) = √(2^2 + 4^2) = √20

Now, let's find the height of the triangle. The height is the perpendicular distance from the point C to AB. To find this distance, we need to calculate the slope of AB and then find the equation of the line perpendicular to AB passing through C.

Slope of AB = (y2 - y1)/(x2 - x1) = (7 - 3)/(4 - 2) = 4/2 = 2

Since the slope of the perpendicular line is the negative reciprocal of the slope of AB, the slope of the perpendicular line passing through C is -1/2.

Using point-slope form, we can write the equation of the perpendicular line passing through C:

y - y1 = m(x - x1), where (x1, y1) = (7, 3) and m = -1/2

y - 3 = (-1/2)(x - 7)
2(y - 3) = -x + 7
2y - 6 = -x + 7
2y + x = 13

Now, we can find the point of intersection of the lines AB and the line perpendicular to AB passing through C. Solving the system of equations:
y = 2x - 1
2y + x = 13

We get x = 3 and y = 5. Therefore, the height of the triangle h = 5.

Now, we can calculate the area using the formula:

Area = (1/2) * base * height
= (1/2) * √20 * 5

Simplifying, we have:
Area = (1/2) * √(4*5) * 5
= (1/2) * √20 * √5
= √5 * √4

Therefore, the area of triangle ABC is √20 units^2, which is option B.