Question

Find the equation of the line passing through \((1,1)\) and forming an area of 2 sq.units with positive coordinate axis. (1) \(2 x+3 y=5\) (2) \(\mathrm{x}-\mathrm{y}+2=0\) (3) \(x+y-2=0\) (4) \(x-y+1=0\)

Short answer

Answer: \(x + y - 2 = 0\)

Step-by-step solution

Check if the lines pass through the point (1, 1)

We will plug in the coordinates (1, 1) into each of the line equations, one by one, and see if they hold true. If the equations hold true, then the line passes through the point (1, 1). (1) \(2 x + 3 y = 5\) Plugging x = 1 and y = 1, we get: \(2(1) + 3(1) = 5\) \(2 + 3 = 5\) \(5 = 5\) Equation (1) is true. (2) \(x - y + 2 = 0\) Plugging x = 1 and y = 1, we get: \(1 - 1 + 2 = 0\) \(2 - 1 = 0\) \(1 \neq 0\) Equation (2) is not true. (3) \(x + y - 2 = 0\) Plugging x = 1 and y = 1, we get: \(1 + 1 - 2 = 0\) \(2 - 2 = 0\) \(0 = 0\) Equation (3) is true. (4) \(x - y + 1 = 0\) Plugging x = 1 and y = 1, we get: \(1 - 1 + 1 = 0\) \(1 - 1 = 0\) \(0 = 0\) Equation (4) is true. So, equations (1), (3), and (4) pass through the point (1, 1).

Calculate the area of the triangle formed by the lines and the coordinate axis

Now, we will calculate the area of the triangle formed by each of these lines and the positive coordinate axis. To do this, we will find the points where each line intersects the x-axis and y-axis. These points, along with the origin, will form the vertices of the triangle. For line (1): \(2x+3y=5\) When x = 0 (intersecting y-axis): \(2(0) + 3y = 5\) \(3y = 5\) \(y = \frac{5}{3}\) When y = 0 (intersecting x-axis): \(2x + 3(0) = 5\) \(2x = 5\) \(x = \frac{5}{2}\) Triangle vertices: \((0, \frac{5}{3})\), \((\frac{5}{2}, 0)\), and \((0, 0)\) Area: \(A_1 = \frac{1}{2} \cdot (\frac{5}{3}) \cdot (\frac{5}{2}) = \frac{25}{12}\) For line (3): \(x+y-2=0\) When x = 0 (intersecting y-axis): \(0 + y - 2 = 0\) \(y = 2\) When y = 0 (intersecting x-axis): \(x + 0 - 2 = 0\) \(x = 2\) Triangle vertices: \((0, 2)\), \((2, 0)\), and \((0, 0)\) Area: \(A_2 = \frac{1}{2} \cdot 2 \cdot 2 = 2\) For line (4): \(x - y + 1 = 0\) When x = 0 (intersecting y-axis): \(0 - y + 1 = 0\) \(y = 1\) When y = 0 (intersecting x-axis): \(x - 0 + 1 = 0\) \(x = -1\) This line does not intersect the positive coordinate axis, so we can discard this line. Now, we have two areas: \(A_1 = \frac{25}{12}\), and \(A_2 = 2\). We need an area of 2 square units, which is only satisfied by the triangle formed by the line (3) \(x+y-2=0\). Therefore, this is the equation of the required line.

Key concepts

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, involves using algebraic methods to solve geometric problems. It connects algebra and geometry through graphs of equations and uses coordinated points on a plane.
Here, we typically discuss points, lines, and distances, while integrating algebraic principles to simplify complex geometric concepts.

• Points are represented as pairs of numbers, such as \(x, y\).
• Lines are described using linear equations.
• The Cartesian coordinate system is often used, consisting of horizontal x-axis and vertical y-axis.

This field allows us to easily find distances between points, midpoints of line segments, and equations of lines. It's a powerful tool for solving geometric problems that helps to visualize and analyze shapes.

Area of Triangle

The area of a triangle formed by three vertices can be calculated using different methods. When dealing with coordinate geometry, it's often simplest to use the formula based on the vertex coordinates.
For a triangle with vertices \(x_1, y_1\), \(x_2, y_2\), and \(x_3, y_3\), the area \(A\) can be calculated as follows:\[ A = \frac{1}{2} | x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) | \] This formula is a practical tool when you have the coordinates and need to find the area quickly.

• The absolute value ensures the area is always positive.
• It provides a straightforward way to deal with triangles in the coordinate plane.

Understanding how to calculate these areas is essential for solving many problems in geometry, especially those involving intersections and line equations.

Point-Slope Form

The point-slope form of a line's equation is extremely useful. It's an effective way to write the equation when you know a point on the line and the slope.
The formula is expressed as:\[ y - y_1 = m(x - x_1) \]where \(m\) is the slope and \(x_1, y_1\) is a known point on the line.

• This form emphasizes the line's slope, making it easy to see how steep the line is.
• It's particularly helpful in deriving equations quickly from given data.

By applying the point-slope form, finding equations of lines becomes systematic and straightforward, which is crucial when grappling with more complex problems, such as finding intersections or calculating areas related to lines.

Intersection with Axes

The intersection points of a line with the axes are crucial for various geometric calculations, such as determining the area of triangles formed with the axes. To find these intersections:

• Y-axis Intersection: Set \(x = 0\) in the line equation, solving for \(y\).
• X-axis Intersection: Set \(y = 0\) in the line equation, solving for \(x\).

These points, together with the origin (0,0), typically form a right-angled triangle with the axes. Understanding these intersection points helps to visualize where lines are located in the plane and how they interact with the coordinate axes.
This makes it easier to solve problems involving calculations of areas and distances on the graph, and to visualize the geometric relationships in a structured manner.