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# Find the area of the triang AOC in the coordinate plane

Question:

In the figure shown, the line L has the equation y = k x + b. The line L pass point C and point B. The point B has the coordinate (-4, 6) and the point C has the coordinate (0, 3). The line L intersects with the x-axis at the point A. What is the area of the triangle AOC in the coordinate plane? Solution:

The area of the triangle AOC is half of the product of base and the altitude to the base. Base is segment OA. Draw a line from the vertex C and perpendicular to the x-axis, this line intersects the x-axis at point D. so the altitude to the base is the segment CD. Area of triangle AOC = (1/2) b × h = (1/2) OA × h = (1/2) OA × CD

Because point C has the coordinate (-4, 6). The y-coordinate of the point C is 6. So, h = CD = 6

We need to find the length of the segment OA that is the x-coordinate of the point A.

Because point B and point C lie on the line L, so point B and point C satisfy the line equation.

Substitute the coordinate of the point C into the line equation, we get

6 = k (-4) + b ... name this as equation1

Substitute the coordinate of the point B into the line equation, we get

3 = k (0) + b ... name this as equation2

From equation2, we get b = 3. Substitute b = 3 into equation1,

6 = -4k + 3

Move the variable term -4k to the left side of the equation and move the constant term 6 to the right side of the equation.

4k = 3 - 6 = -3
k = -3/4, the slope of the line L is -3/4
The equation of the line L is y = -(3/4) x + 3

Because the line L intersects the x-axis at point A, so the point A has the coordinate (x, 0).

Substitute the coordinate of the point A into the line equation, we get

0 = -(3/4) x + 3

Move the variable term into the left side of the equation

(3/4) x = 3

Both side of the equation divide by 3

(1/4) x = 1

Both side of the equation times 4

x = 4
OA = 4
The area of the triangle AOC = (1/2) × OA × CD = (1/2) × 4 × 6 = 12

Therefore, the area of the triangle AOC is 12. Please watch the video for more details.