What is the area (in unit squares) of the triangle enclosed by the graphs of 2x + 5y = 12, x + y = 3 and the x-axis?
3
The problem asks for the area of the triangle enclosed by the graphs of three lines: 2x + 5y = 12, x + y = 3, and the x-axis. The x-axis is represented by the equation y = 0.
To find the area of the triangle, we first need to identify its vertices. The vertices are the points where these three lines intersect.
There are three vertices, corresponding to the intersections of the three pairs of lines:
Substitute $y = 0$ into the first equation:
$$2x + 5(0) = 12$$ $$2x = 12$$ $$x = \frac{12}{2} = 6$$
So, Vertex 1 is (6, 0).
Substitute $y = 0$ into the second equation:
$$x + 0 = 3$$ $$x = 3$$
So, Vertex 2 is (3, 0).
We have a system of two linear equations:
From Equation 2, we can express x in terms of y:
$$x = 3 - y$$
Now, substitute this expression for x into Equation 1:
$$2(3 - y) + 5y = 12$$ $$6 - 2y + 5y = 12$$ $$6 + 3y = 12$$ $$3y = 12 - 6$$ $$3y = 6$$ $$y = \frac{6}{3} = 2$$
Now, substitute the value of y back into the expression for x:
$$x = 3 - y = 3 - 2 = 1$$
So, Vertex 3 is (1, 2).
The three vertices of the triangle are (6, 0), (3, 0), and (1, 2).
Two of the vertices, (6, 0) and (3, 0), lie on the x-axis. This segment on the x-axis forms the base of the triangle. The length of the base is the distance between these two points:
$$Base = |6 - 3| = 3 \text{ units}$$
The height of the triangle is the perpendicular distance from the third vertex, (1, 2), to the base (the x-axis). This distance is the absolute value of the y-coordinate of the third vertex.
$$Height = |2| = 2 \text{ units}$$
The area of a triangle is given by the formula:
$$Area = \frac{1}{2} \times Base \times Height$$
Using the base and height we found:
$$Area = \frac{1}{2} \times 3 \times 2$$ $$Area = \frac{1}{2} \times 6$$ $$Area = 3 \text{ square units}$$
The area of the triangle enclosed by the given lines and the x-axis is 3 unit squares.
| Line Equation | Description |
|---|---|
| $2x + 5y = 12$ | A linear equation |
| $x + y = 3$ | Another linear equation |
| y = 0 | The x-axis |
| Concept | Details |
|---|---|
| Vertices of the triangle | (6, 0), (3, 0), (1, 2) |
| Base of the triangle | Segment on x-axis between (3,0) and (6,0). Length = 3. |
| Height of the triangle | Perpendicular distance from (1,2) to x-axis. Height = 2. |
| Area Formula | $\frac{1}{2} \times Base \times Height$ |
| Calculated Area | 3 square units |
Lines in a coordinate plane can intersect to form geometric shapes like triangles. Finding the vertices is crucial for calculating the area or other properties of the shape. When one of the bounding lines is an axis (like the x-axis, y=0), it often simplifies finding the base or height of the triangle.
Understanding how to solve systems of linear equations and basic coordinate geometry concepts is fundamental to solving this type of problem.
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