Class 11 RS Aggarwal Chapter-20 Straight Lines Solutions - Free PDF Download
FAQs on RS Aggarwal Class 11 Solutions Chapter-20 Straight Lines
1. What is the correct step-by-step method to find the equation of a line when two points are given, as per the RS Aggarwal Class 11 solutions for Chapter 20?
To find the equation of a line passing through two given points, (x₁, y₁) and (x₂, y₂), you should follow the two-point form methodology demonstrated in the solutions. The steps are:
- First, calculate the slope (m) of the line using the formula: m = (y₂ - y₁) / (x₂ - x₁).
- Next, substitute the calculated slope (m) and one of the given points (either (x₁, y₁) or (x₂, y₂)) into the point-slope form: y - y₁ = m(x - x₁).
- Finally, simplify the equation into the general form Ax + By + C = 0 as required by the specific problem.
2. How do the RS Aggarwal solutions for Chapter 20 suggest using the intercept form of a line to solve problems?
The RS Aggarwal solutions utilise the intercept form of a line, which is x/a + y/b = 1, primarily when a problem provides the x-intercept (a) and the y-intercept (b). The correct method is to directly substitute the values of 'a' and 'b' into this formula. This approach is the most efficient way to find the equation when the points where the line crosses the coordinate axes are explicitly known.
3. What is the standard procedure to calculate the perpendicular distance from a point to a line, following the method in RS Aggarwal solutions?
The standard procedure to find the perpendicular distance from a point (x₁, y₁) to a line is to first ensure the line's equation is in the general form (Ax + By + C = 0). Then, apply the distance formula: d = |Ax₁ + By₁ + C| / √(A² + B²). The solutions emphasise substituting the coordinates of the point and the coefficients from the line's equation correctly to get the accurate distance.
4. When solving problems from RS Aggarwal Chapter 20, how do I decide which form of a linear equation is the most efficient to use?
Choosing the most efficient form of a linear equation depends on the information given in the problem. A good strategy is as follows:
- Use point-slope form (y - y₁ = m(x - x₁)) when you know one point and the slope.
- Use two-point form when you are given two points on the line.
- Use slope-intercept form (y = mx + c) when the slope and the y-intercept are known.
- Use intercept form (x/a + y/b = 1) when both the x- and y-intercepts are provided.
- Use normal form (x cos α + y sin α = p) for problems involving the perpendicular distance from the origin.
5. Why is it necessary to convert the general equation of a line into its normal form to determine the perpendicular distance from the origin?
Converting the general equation Ax + By + C = 0 to the normal form, x cos α + y sin α = p, is crucial because the value 'p' in the normal form directly represents the length of the perpendicular from the origin to the line. The general form does not provide this distance explicitly. The conversion process standardises the equation, and 'p' is always positive, correctly representing a physical distance.
6. What is a common mistake when finding the angle between two lines, and how do the RS Aggarwal solutions help in avoiding it?
A common mistake is incorrectly identifying the angle required. The formula tan θ = |(m₂ - m₁) / (1 + m₁m₂)| gives the acute angle (θ) between the two lines. If the problem requires the obtuse angle, students often forget to calculate it as 180° - θ. The RS Aggarwal solutions typically clarify through diagrams and step-by-step explanations whether the acute or obtuse angle is the intended answer based on the problem's context.
7. How are the concepts of parallel and perpendicular lines used to solve problems for finding the equation of a new line in RS Aggarwal Chapter 20?
These concepts are fundamental for finding the slope of the required line.
- If a line is parallel to a given line, their slopes are equal (m₁ = m₂). You use the slope of the given line for your new equation.
- If a line is perpendicular to a given line, the product of their slopes is -1 (m₁ * m₂ = -1). You calculate the negative reciprocal of the given line's slope to find the slope of your new line.
