Question

Ihe equatron of the line making an angle of \(45^{\circ}\) with X-axis \(1 n\) positive direction and having y-intercept as \(-3\) is (1) \(3 \mathrm{x}-\mathrm{y}+1=0\) (2) \(3 \mathrm{x}+\mathrm{y}-1=0\) (3) \(x-y+3=0\) (4) \(x-y=3\)

Short answer

Based on the given information, the correct equation for the line is \(x - y - 3 = 0\), and the correct option is (3).

Step-by-step solution

Calculate the slope of the line

The angle between the line and the positive X-axis is \(45^{\circ}\). We can find the slope (m) of the line using the tangent of the angle: \(m = tan(45^{\circ})\). Since \(tan(45^{\circ}) = 1\), the slope of our line is: \(m = 1\)

Write the equation of the line using the point-slope form

We are given the y-intercept: \(y = -3\) when \(x = 0\). So, the line passes through point (0, -3). Using the point-slope form and the slope we found in step 1: \(y - y_{1} = m(x - x_{1})\) Substitute the values of the point and the slope: \(y - (-3) = 1 \cdot (x - 0)\) Simplify the equation: \(y + 3 = x\)

Compare the obtained equation with the given options

Now, we need to rewrite the equation in the form used in the options and compare it to the given options: \(x - y -3 = 0\) Comparing the equation with the given options, we find that it matches with option (3): \(x - y + 3 = 0\) So, the correct answer is option (3).

Key concepts

Understanding Linear Equations

Linear equations are foundational in algebra and describe relationships where the output is directly proportional to the input. They take the general form of \(ax + by + c = 0\), where \(a\), \(b\), and \(c\) are constants. In simple terms, a linear equation will always graph a straight line on the coordinate plane.
It is important to recognize that the coefficients \(a\) and \(b\) determine the direction and slope of the line, while the constant \(c\) adjusts its position. For instance, in our exercise, the equation \(x - y + 3 = 0\) represents a line where every point \(x, y\) satisfies this relationship.
To solve or manipulate linear equations, one might rearrange them, substitute values for known variables, or use algebraic techniques to find intersections with axes or other lines. This flexibility makes linear equations immensely powerful in modeling and solving real-world problems.

Exploring Coordinate Geometry

Coordinate geometry, also known as analytic geometry, combines algebra and geometry to study the position of points, lines, and shapes on the Cartesian plane. It uses numerical coordinates to describe the position of points and can precisely define the properties of geometric figures.
In the provided exercise, the coordinate plane is crucial for graphing the linear equation. We plot points, like the y-intercept \((0, -3)\), to represent the line graphically.

• Coordinates: Pairs like \((x, y)\) provide the location of points.
• Axes: The x-axis and y-axis are perpendicular, dividing the plane into four quadrants.

The coordinate plane allows us to visualize equations and understand their geometric representation, making coordinate geometry a vital tool in mathematics education.

Concept of Angle with X-Axis

Angles in coordinate geometry help describe the orientation of lines. The angle made with the x-axis provides insight into the line's slope. For a line making an angle \(\theta\) with the positive x-axis, the slope \(m\) is determined by \(m = \tan(\theta)\).
In our problem, the line forms a \(45^{\circ}\) angle with the x-axis, which gives it a slope of \(1\) because \(\tan(45^{\circ}) = 1\). This slope means for every unit increase in \(x\), \(y\) increases by an equal amount.
Understanding this concept helps us quickly identify lines that rise or fall in specific ways, crucial for graph analysis and predicting linear behaviors.

Point-Slope Form of a Line

The point-slope form is a method to write the equation of a line when you know a point on the line and its slope. The formula is expressed as \(y - y_1 = m(x - x_1)\), where \((x_1, y_1)\) is a known point and \(m\) is the slope.
This form is particularly useful because it directly uses these known quantities to form the line's equation. In the exercise, the slope \(m = 1\) and point \((0, -3)\) are used to form the equation \(y - (-3) = 1 \cdot (x - 0)\), simplifying to \(y + 3 = x\).
This approach simplifies transitioning from conceptual understanding to practical equation use, making it easier to tackle problems involving lines in coordinate geometry.