Step 1: Find the intercepts.
Given equation: \[ x + 2y = 4 \] To find the X-intercept, put \(y = 0\): \[ x = 4 \] Hence, the X-intercept is \((4, 0)\). To find the Y-intercept, put \(x = 0\): \[ 2y = 4 \implies y = 2 \] Hence, the Y-intercept is \((0, 2)\).
Step 2: Plot the graph.
Plot the points \((4, 0)\) and \((0, 2)\) on a graph paper and draw a straight line joining them. This line intersects the X-axis at \((4, 0)\) and the Y-axis at \((0, 2)\).
Step 3: Find the area of the triangle.
The line forms a right-angled triangle with the coordinate axes. Base = 4 units, Height = 2 units. \[ \text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height} \] \[ \text{Area} = \frac{1}{2} \times 4 \times 2 = 4 \text{ sq. units.} \] Step 4: Conclusion.
Hence, the area of the triangle formed by the line and the coordinate axes is \(4\) square units.
Final Answer: \[ \boxed{\text{Area = 4 square units}} \]
Let $ \mathbb{R} $ denote the set of all real numbers. Then the area of the region $$ \left\{ (x, y) \in \mathbb{R} \times \mathbb{R} : x > 0, y > \frac{1}{x},\ 5x - 4y - 1 > 0,\ 4x + 4y - 17 < 0 \right\} $$ is
Let \( C_{t-1} = 28, C_t = 56 \) and \( C_{t+1} = 70 \). Let \( A(4 \cos t, 4 \sin t), B(2 \sin t, -2 \cos t) \text{ and } C(3r - n_1, r^2 - n - 1) \) be the vertices of a triangle ABC, where \( t \) is a parameter. If \( (3x - 1)^2 + (3y)^2 = \alpha \) is the locus of the centroid of triangle ABC, then \( \alpha \) equals: