Final answer:
The correct answer is option B. 50 square units. The area of the triangle formed by the y-axis, the line f(x) = 10 - 1/4x, and the line perpendicular to f(x) that passes through the origin is 50 square units.
Step-by-step explanation:
To find the area of the triangle, we need to find the coordinates of the points where the lines intersect with the y-axis and then use the formula for the area of a triangle.
The line perpendicular to f(x) passes through the origin, so it has the equation g(x) = -4x.
The intersection points with the y-axis can be found by substituting x=0 into each equation. For f(x), we have f(0) = 10. For g(x), we have g(0) = 0.
Now, we can calculate the area of the triangle using the formula: Area = 1/2 * base * height. The base is the distance between the intersection points on the y-axis, which is 10 units. The height is the perpendicular distance from the line g(x) to the y-axis, which is also 10 units.
Therefore, the area of the triangle is: Area = 1/2 * 10 * 10 = 50 square units.
To find the area of the triangle, we first need to find the points where the line f(x) = 10 - \frac{1}{4}x intersects the y-axis and where the line perpendicular to f(x) that passes through the origin intersects the y-axis. The line f(x) intersects the y-axis when x = 0, which gives us the point (0, 10).
The perpendicular line will have a slope that is the negative reciprocal of the slope of f(x). Since the slope of f(x) is -\frac{1}{4}, the slope of the perpendicular line is 4. The equation of the perpendicular line passing through the origin (0,0) is y = 4x. This line intersects the y-axis when x = 0, which is also at the point (0, 0).
Therefore, the triangle formed has a base of 10 units along the y-axis (from the point (0, 0) to the point (0, 10)) and a height which is the x-coordinate of the point where f(x) intersects the y-axis, which is also 10 units. The area of a triangle is given by A = \frac{1}{2} \times base \times height, which in this case is \frac{1}{2} \times 10 \times 10 = 50 square units.