Triangle ABC is graphed on the coordinate plane. Determine the area of triangle ABC using BC as the base

Question

A coordinates (2,5) 

B coordinates (10,9)

C coordinates (6,1)

Answer 1

If you are familiar with the distance formula, D = [(x₂-x₁)²+(y₂-y₁)²]^1/2, then it's a matter of plugging values:

 

NOTE: if not apparent, the 1/2 exponent means square root.

 

For BC: Let x₁= 6, x₂= 10, y₁ = 1, and y₂ = 9.

Then the distance, or length of the base BC is:

D = [(10-6)² + (9-1)²]^1/2

D = [(4)² + (8)²]^1/2

D = [16+64]^1/2

D = [80]^1/2

D = 4(5)^(1/2)

 

So, the length of BC is 4(5)^(1/2) units.

 

Now for AC: Let x₁= 2, x₂= 6, y₁ = 5, and y₂ = 1.

The distance of AC is:

D = [(6-2)² + (1-5)²]^1/2
D = [(4)_² + (-4)²]^1/2
D = [16+16]^1/2
D = [32]^1/2
D = 4(2)^1/2 (when you pull out a 16 from the square root you take that root and get 4 with a remainder of 2 under the root)

 

So, the length of AC is 4(2)^1/2

 

The area of a triangle is 1/2*base*height. The base, BC was 4(5)^(1/2) and the height AC was 4(2)^1/2. Thus the area of triangle ABC is:

 

A = 1/2*(4(5)^1/2)*[4(2)^1/2]

A = 8(10)^1/2 units.

 

The solution is 8(10)^1/2 units (that is 8 times the square root of 10 units).

 

 

Answer 2

I like to get my bearings in a question like this by sketching the triangle, and figuring out something about its sides. Plotting the three points, it looks as if the vector from A to C would be (4,-4), the one from A to B would be (8,4), and the vector from C to B would be (4,8). You should check these for yourself, on your own sketch of the triangle.

 

There are several triangle area formulas we could use, and it sounds like you want to work with A=(1/2)*b*h. If BC is the base, then we'll need its length: sqrt(4*4+8*8)=4*sqrt(5). Now we need to know the height, that is, the length of a perpendicular line from point A to line BC.

 

Since BC has a slope of 2, anything perpendicular to it must have a slope of -1/2. A line through (2,5) with a slope of -1/2 is given by the equation x+2y=12. The line BC is given by the equation 2x-y=11. These two lines intersect at the point (6.8,2.6) exactly.

Now we can find the height of the triangle from base BC: it's the distance from (2,5) to (6.8,2.6). The distance formula gives us sqrt(4.8*4.8 + 2.4*2.4) = 12*sqrt(5)/5.

 

Now it's straightforward to calculate: A=(1/2)*b*h = (1/2)*4sqrt(5)*12sqrt(5)/5, which you can simplify.

 

Does that help?