Straight Lines Class 11 One Shot | NCERT Maths Complete Chapter-9 Important Questions | CBSE 2025
Next Toppers - 11th Science・17 minutes read
Understanding coordinate geometry is essential for competitive exams, focusing on concepts like slopes, distances, and line equations. Students should start with foundational NCERT textbooks and progress to advanced practices to master the relationships between lines, ensuring they grasp key formulas and methods for various geometric applications.
Insights
- Competitive exams heavily feature coordinate geometry, especially the chapter on straight lines, which is essential for mastering higher-level mathematics in grades 11 and 12.
- The slope of a line can be determined from the angle it makes with the x-axis, and the relationship between the slopes of two lines can be analyzed using the tangent of the angle between them.
- When the tangent of the angle between two lines is 1/3, it indicates that one line's slope is double that of the other, facilitating the calculation of both slopes.
- The principle that the product of the slopes of two perpendicular lines is -1 is crucial for solving problems related to perpendicular relationships in coordinate geometry.
- The equation of a line can be derived from two points using the linear equation form, such as 2x + 3y = 4, which ensures all points on the line satisfy this equation.
- The distance formula for calculating the distance between two points in a 2D plane is fundamental in geometry and is expressed as √((x2 - x1)² + (y2 - y1)²).
- The section formula is a method for finding the coordinates of a point that divides a line segment in a specific ratio, expressed mathematically for both x and y coordinates.
- The midpoint formula, which finds the midpoint of a line segment, is useful for dividing segments evenly and is calculated as ((x1 + x2)/2, (y1 + y2)/2).
- The area of a triangle formed by three points can be calculated using a specific formula, providing a straightforward method for determining area based on vertex coordinates.
- For effective exam preparation in coordinate geometry, students should begin with foundational NCERT textbooks before advancing to more complex reference materials.
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Recent questions
What is the slope of a line?
The slope of a line is a measure of its steepness, calculated as the ratio of the vertical change (rise) to the horizontal change (run) between two points on the line. It is often represented by the letter "m" and can be determined using the formula \( m = \frac{y_2 - y_1}{x_2 - x_1} \), where \( (x_1, y_1) \) and \( (x_2, y_2) \) are the coordinates of two distinct points on the line. A positive slope indicates that the line rises as it moves from left to right, while a negative slope indicates that it falls. A slope of zero represents a horizontal line, and an undefined slope corresponds to a vertical line.
How to find the distance between two points?
The distance between two points in a 2D plane can be calculated using the distance formula, which is derived from the Pythagorean theorem. If you have two points, \( (x_1, y_1) \) and \( (x_2, y_2) \), the distance \( d \) between them is given by the formula \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \). This formula allows you to find the straight-line distance between the two points by taking the square root of the sum of the squares of the differences in their x-coordinates and y-coordinates. It is a fundamental concept in geometry and is widely used in various applications, including navigation and computer graphics.
What is the midpoint formula?
The midpoint formula is used to find the coordinates of the midpoint of a line segment defined by two endpoints. If the endpoints are given as \( (x_1, y_1) \) and \( (x_2, y_2) \), the midpoint \( M \) can be calculated using the formula \( M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \). This formula averages the x-coordinates and the y-coordinates of the two endpoints, providing the exact center point of the segment. The midpoint is particularly useful in various fields, including geometry, computer graphics, and physics, as it helps in determining the center of a line segment or in dividing it into equal parts.
What is the equation of a line?
The equation of a line can be expressed in several forms, with the most common being the slope-intercept form, which is written as \( y = mx + b \). In this equation, \( m \) represents the slope of the line, indicating its steepness, and \( b \) is the y-intercept, the point where the line crosses the y-axis. Another common form is the point-slope form, given by \( y - y_1 = m(x - x_1) \), where \( (x_1, y_1) \) is a specific point on the line. Additionally, the standard form of a line is expressed as \( Ax + By = C \), where \( A \), \( B \), and \( C \) are constants. Each of these forms provides a different perspective on the line's characteristics and can be used depending on the context of the problem.
How to determine if two lines are parallel?
To determine if two lines are parallel, you need to compare their slopes. If the slopes of the two lines are equal, then the lines are parallel and will never intersect. For example, if you have two lines represented by their equations in slope-intercept form \( y = m_1x + b_1 \) and \( y = m_2x + b_2 \), you can simply check if \( m_1 = m_2 \). If they are equal, the lines are parallel. Additionally, if the lines are given in standard form \( Ax + By = C \), you can rearrange them to find their slopes or compare the ratios of their coefficients. Parallel lines have the same direction and maintain a constant distance apart, which is a fundamental concept in geometry and is applicable in various real-world scenarios, such as architecture and engineering.
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Summary
00:00
Mastering Coordinate Geometry for Competitive Exams
- Competitive exams often include concepts from coordinate geometry, particularly the chapter on straight lines, which is crucial for understanding higher-level mathematics in grades 11 and 12.
- The slope of a line can be derived from the angle theta formed with the x-axis, and the relationship between slopes of two lines can be determined using the tangent of the angle between them.
- If the tangent of the angle between two lines is 1/3, the slope of one line is double that of the other, allowing for the calculation of their respective slopes.
- The product of the slopes of two perpendicular lines is always -1, which is essential for solving problems involving perpendicular relationships in coordinate geometry.
- To find the equation of a line given two points (x1, y1) and (x2, y2), the linear equation can be expressed in the form 2x + 3y = 4, where all points on the line satisfy this equation.
- The distance between two points in a 2D plane can be calculated using the formula √((x2 - x1)² + (y2 - y1)²), which is fundamental in geometry.
- The section formula allows for finding the coordinates of a point dividing a line segment in a given ratio, expressed as (mx2 + nx1)/(m+n) for the x-coordinate and (my2 + ny1)/(m+n) for the y-coordinate.
- The midpoint formula, applicable when the ratio is 1:1, is given by ((x1 + x2)/2, (y1 + y2)/2), providing the coordinates of the midpoint of a line segment.
- The area of a triangle formed by three points (x1, y1), (x2, y2), and (x3, y3) can be calculated using the formula: Area = 1/2 |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|.
- For effective exam preparation, students are advised to start with NCERT textbooks for foundational practice before progressing to reference books for advanced concepts in coordinate geometry.
14:14
Finding Ratios and Slopes of Lines
- Given coordinates are -11 for point A and unspecified for point B; a line is formed by connecting these points.
- The equation of a line is provided as x + y = 4, which intersects at point P.
- To find the ratio in which point P divides the line, let the ratio be k:1, with coordinates of P as (α, β).
- The formula for α is derived as (5k - 1) / (k + 1) and for β as (7k + 1) / (k + 1).
- Substituting α and β into the line equation x + y = 4 leads to the equation 12k = 4k + 4.
- Solving for k gives k = 1/2, indicating point P divides the line in the ratio 1:2.
- The inclination of a line is defined as the angle θ formed with the positive x-axis, measured anticlockwise.
- The slope (m) of a line is calculated as m = tan(θ), indicating the line's steepness.
- The slope can be calculated using two points (x1, y1) and (x2, y2) with the formula m = (y2 - y1) / (x2 - x1).
- A horizontal line has a slope of 0, while a vertical line's slope is considered undefined.
28:07
Understanding Slopes and Line Relationships
- The slope of vertical lines is undefined, while horizontal lines have a slope of zero, indicating their distinct orientations in a coordinate system.
- For two parallel lines, their slopes (m1 and m2) are equal, meaning if m1 = m2, the lines do not intersect.
- To find the slope between two points (x1, y1) and (x2, y2), use the formula: slope = (y2 - y1) / (x2 - x1).
- For perpendicular lines, the product of their slopes must equal -1, expressed as m1 * m2 = -1, indicating a 90-degree angle between them.
- When calculating slopes, ensure to simplify expressions correctly; for example, if y2 - y1 = 5 - (-1) and x2 - x1 = 2 - 0, the slope becomes 6/2 = 3.
- To demonstrate that two lines are parallel, calculate their slopes; if both slopes equal -4/7, the lines are confirmed parallel.
- For perpendicular lines, if one slope is -5/2, the other must be 2/5 to satisfy the condition m1 * m2 = -1.
- The angle between two lines can be calculated using the formula: tan(θ) = |(m2 - m1) / (1 + m1 * m2)|, ensuring the denominator is not zero.
- When given specific points, always draw a figure to visualize the problem, enhancing clarity and understanding of the relationships between lines.
- Practice solving various types of problems, including those from NCERT and miscellaneous sources, to strengthen understanding of line relationships and slopes.
42:04
Calculating Slopes and Angles Between Lines
- The slope of a line can be calculated using the formula \( m = \frac{y_2 - y_1}{x_2 - x_1} \), where \( y_1, y_2 \) and \( x_1, x_2 \) are coordinates of two points on the line.
- For the line with points \( (x_1, y_1) = (-2, -4) \) and \( (x_2, y_2) = (-2, -3) \), the slope \( m_1 \) is calculated as \( \frac{-3 - (-4)}{-2 - (-2)} = \frac{1}{0} \), indicating a vertical line.
- To find the slope \( m_2 \) of another line with points \( (x_1, y_1) = (-2, -2) \) and \( (x_2, y_2) = (1, -3) \), use the same formula: \( m_2 = \frac{-3 - (-2)}{1 - (-2)} = \frac{-1}{3} \).
- The angle \( \theta \) between two lines with slopes \( m_1 \) and \( m_2 \) can be calculated using the formula \( \tan(\theta) = \frac{m_2 - m_1}{1 + m_1 \cdot m_2} \).
- If \( m_1 = \frac{7}{4} \) and \( m_2 = \frac{1}{2} \), then \( \tan(\theta) = \frac{\frac{1}{2} - \frac{7}{4}}{1 + \frac{7}{4} \cdot \frac{1}{2}} \).
- The calculation simplifies to \( \tan(\theta) = \frac{-\frac{5}{4}}{\frac{9}{8}} = -\frac{10}{9} \), indicating the angle between the two lines.
- A problem states that the slope of one line is double that of another, with the tangent of the angle between them given as \( \frac{1}{3} \).
- Setting \( m_1 = 2m_2 \) and using the tangent formula, we derive \( \frac{m_2 - 2m_2}{1 + 2m_2^2} = \frac{1}{3} \).
- Solving the equation leads to two possible values for \( m_2 \): \( \frac{1}{2} \) and \( -\frac{1}{2} \), resulting in corresponding values for \( m_1 \) as \( 1 \) and \( -1 \).
- The equations of lines parallel to the x-axis or y-axis are represented as \( y = k \) or \( x = k \), where \( k \) is a constant, indicating the fixed value of either coordinate.
56:58
Understanding Linear Equations and Their Properties
- The value of x is fixed while y can vary; if y is fixed, the equation becomes y = c or y = k, indicating a constant value for y.
- A line parallel to the y-axis has a constant x value, leading to the equation x = constant, while y can take any value.
- Given two points, the equation of a line can be formed using the point-slope form, which requires the slope and one point's coordinates.
- To find the slope from two points, use the formula m = (y2 - y1) / (x2 - x1); this allows calculation of the line's slope.
- For a point (2, 3) with a slope m = 2, the equation can be derived as y - 3 = 2(x - 2), simplifying to 2x - y - 1 = 0.
- To find the equation of a line making a 135° angle with the x-axis, calculate the slope using tan(135°) = -1, then apply the point-slope formula.
- The two-point form of a line's equation is y - y1 = (y2 - y1) / (x2 - x1) * (x - x1), allowing for the creation of an equation from two given points.
- The slope-intercept form is y = mx + c, where m is the slope and c is the y-intercept; this form is essential for graphing linear equations.
- For a line with slope 1/2 and y-intercept -1/4, the equation becomes y = (1/2)x - 1/4, which can be rearranged to 2x - 4y = 5.
- To prove two lines are parallel, confirm their slopes are equal; if the slopes are the same, the lines do not intersect and are parallel.
01:11:35
Understanding Slope and Line Equations
- To find the slope from the equation y = mx + b, rearrange it into the form y = mx + c, identifying m as the slope and c as the y-intercept.
- For the equation 2y = -x + 9, divide by 2 to get y = -1/2x + 9/2, where the slope m is -1/2 and the y-intercept c is 9/2.
- The slope-intercept form y = mx + c allows for easy identification of the slope (m) and y-intercept (c) from any linear equation.
- To find the angle θ that a line makes with the positive x-axis, use the slope m, where tan(θ) = m; for m = -1/√3, θ = 150° or 5π/6 radians.
- For the two-point form of a line, use the formula (y - y1) / (x - x1) = (y2 - y1) / (x2 - x1) to derive the line's equation from two given points.
- To prove three points are collinear, check if the equation derived from any two points satisfies the third point; if true, they lie on the same line.
- The equation of the line through points (3, 0) and (8, 2) is derived as 2x - 5y - 6 = 0, confirming the relationship between the points.
- The perpendicular bisector of a line segment between two points is both perpendicular to the segment and divides it into two equal parts, requiring specific calculations for its equation.
- To find the slope from the equation x/√3 - y - 6 = 0, rearrange to y = -1/√3x + 6/√3, identifying the slope as -1/√3.
- The slope can be converted into various forms, including point-slope and slope-intercept forms, allowing flexibility in representing linear equations.
01:26:14
Understanding Line Equations and Slopes
- To apply the slope point form, you need to know either a point and the slope or two points to derive the slope of the line.
- The midpoint formula for two points (x1, y1) and (x2, y2) is calculated as: ((x1 + x2)/2, (y1 + y2)/2).
- For points (6, 3) and (2, -5), the midpoint is ((6 + 2)/2, (3 - 5)/2) = (4, -1).
- The slope (m) between two points is calculated as: m = (y2 - y1) / (x2 - x1). For points (6, 3) and (2, -5), m = (-5 - 3) / (2 - 6) = 2.
- If two lines are perpendicular, the product of their slopes (m1 * m2) equals -1. For m1 = 1/2, m2 = -2.
- The equation of a line can be formed using the point-slope form: y - y1 = m(x - x1). For point (4, -1) and slope 1/2, it becomes y + 1 = 1/2(x - 4).
- Rearranging gives the equation: 2y + 2 = x - 4, or x - 2y - 6 = 0.
- The slope-intercept form of a line is y = mx + b, where m is the slope and b is the y-intercept.
- For intercepts a = 2 and b = -3, the equation is x/2 + y/(-3) = 1, which simplifies to 3x + 2y = 6.
- To find the equation of a line with intercepts whose product is -6, use the form x/a + y/b = 1, solving for a and b based on given conditions.
01:40:43
Finding Line Equations with Intercepts
- The task involves finding the equation of a line that passes through a specific point and has intercepts on the axes, using the intercept form of the equation.
- The intercept form is expressed as \( \frac{x}{a} + \frac{y}{b} = 1 \), where \( a \) and \( b \) represent the x-intercept and y-intercept, respectively.
- Given the point (-3, 7), substitute into the intercept form to derive the equation, leading to \( -x + y = a \).
- To find the value of \( a \), substitute \( x = -3 \) and \( y = 7 \) into the equation, resulting in \( a = 10 \).
- The final equation of the line can be expressed as \( x - y + 10 = 0 \), which simplifies to \( y = x + 10 \).
- A new problem involves finding the equation of a line that divides the segment between the axes in a ratio of 1:2, requiring knowledge of the coordinates of the intercepts.
- The coordinates of the intercepts are represented as \( (0, b) \) and \( (a, 0) \), with the point dividing the segment being \( R \) in the ratio 1:2.
- Using the section formula, the coordinates of point \( R \) can be calculated as \( \left(\frac{2a}{3}, \frac{b}{3}\right) \).
- The equation of the line can be derived using the intercept form, substituting the values of \( a \) and \( b \) to yield \( 2x + 3y = 3hk \).
- The shortest distance from a point to a line can be calculated using the formula \( \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}} \), where \( (x_1, y_1) \) is the point and \( ax + by + c = 0 \) is the line equation.
01:56:09
Calculating Distances to Parallel Lines
- The equations 2x + 3y + 4 = 0 and 4x + 6y - 5 = 0 represent two parallel lines, as their coefficients are proportional.
- To find the distance between a point and two parallel lines, first ensure the equations have equal coefficients by multiplying the first equation by 2.
- The modified first equation becomes 4x + 6y + 8 = 0, allowing for the comparison with the second equation.
- To calculate the distance from the point (-1, 1) to the line 12x - 5y + 82 = 0, use the distance formula: |ax1 + by1 + c| / √(a² + b²).
- For the line 12x - 5y + 82 = 0, substitute a = 12, b = -5, x1 = -1, and y1 = 1 into the formula to find the distance.
- The calculation yields a distance of 65 / √169, simplifying to 5 units from the point to the line.
- To find the perpendicular distance from the origin to the line 4x + 3y - 2 = 0, apply the distance formula with a = 4, b = 3, and c = -2.
- The distance from the origin to the line is calculated as |0 + 0 - 2| / √(4² + 3²), resulting in a distance of 2/5 units.
- The relationship between the lengths of perpendiculars from the origin to two lines can be expressed as p1² = 4p2², where p1 and p2 are the distances to the respective lines.
- The distance formula can be applied to find points on the y-axis with a specific perpendicular distance from a line, such as 3 units from the line 4x - 3y = 12.
02:11:44
Solving Line Equations and Distances Explained
- The equation manipulation begins with removing the third model, leading to the expression -3y + 12, which can be simplified and written multiple times for clarity.
- When solving for y, the equation 3y = 12/5 is derived, requiring consideration of both positive and negative cases for accurate results.
- The positive case yields y = 1, while the negative case results in y = -9, establishing two potential coordinates: (0, 1) and (0, -9).
- The distance between two parallel lines is calculated using the formula: distance = |c1 - c2| / √(a² + b²), where c1 and c2 are constants from the line equations.
- For the lines represented as -3x + 4y - 31 = 0 and another line, the distance is computed as |c1 - c2|, with specific values substituted for accurate results.
- The distance formula involves calculating the square root of the sum of squares of coefficients a and b, ensuring both lines are in standard form for comparison.
- The introduction covers essential concepts like the distance formula, section formula, area of triangles, and slope calculations, crucial for understanding line equations.
- The slope of parallel lines is equal, while the product of slopes of perpendicular lines equals -1, providing foundational knowledge for line relationships.
- The intercept form of a line is derived from the equation 3x + 2y - 12 = 0, transforming it into x/4 + y/6 = 1 to identify x and y intercepts.
- The intersection of three lines is determined by solving their equations simultaneously, leading to coordinates that satisfy all three line equations, demonstrating practical application of algebraic methods.
02:25:59
Angle Relationships and Line Calculations Explained
- The angle \(90 - r\) indicates that if angles \(i\) and \(r\) are equal, then \(90 - i\) and \(90 - r\) will also be equal, establishing a relationship between these angles.
- By defining \(90 - aa\) and \(90 - r\) as \(\theta\), the value of \(\theta\) can be assumed for further calculations involving the angles in the context.
- The slope of a line can be calculated using the formula \(m = \frac{y_2 - y_1}{x_2 - x_1}\), where \(y_2 = 3\), \(y_1 = 0\), \(x_2 = 5\), and \(x_1 = x\).
- The slope of the line is determined to be \(m = \frac{3 - 0}{5 - x}\), leading to the equation \(5 - x = 2\) after simplification.
- Solving the equation \(5 - x = 2\) results in \(x = \frac{13}{5}\), providing the coordinate on the x-axis.
- The equations of the lines given are \(y = 3x + 1\) and \(2y = x + 3\), which can be simplified to \(y = \frac{1}{2}x + \frac{3}{2}\) for further analysis.
- To find the slopes of the lines, the slope of \(y = 3x + 1\) is \(3\) and for \(y = \frac{1}{2}x + \frac{3}{2}\) it is \(\frac{1}{2}\).
- The angles between the lines can be calculated using the formula \(\tan(\theta) = \frac{m_2 - m_1}{1 + m_1m_2}\), where \(m_1\) and \(m_2\) are the slopes of the lines.
- The image of a point with respect to a line can be found by ensuring the distances from the point to the line and from the image to the line are equal, using the line equation \(x + 3y = 7\).
- The coordinates of the image point can be derived by solving the equations formed from the distances, leading to the final coordinates based on the original point and the line equation.
02:40:46
Calculating Slopes and Intersections of Lines
- The slope of the line represented by the equation \(3y = -x + 7\) is calculated as \(-\frac{1}{3}\) after rearranging to \(y = -\frac{1}{3}x + \frac{7}{3}\).
- For two lines to be perpendicular, the product of their slopes must equal \(-1\). If \(m_1 = -1\), then \(m_2\) must equal \(3\).
- The equation \(3h - k = 1\) is derived from manipulating the slope and intercepts, leading to the relationship between \(h\) and \(k\).
- The second equation \(3k + h = -13\) is established, allowing for simultaneous equations to be solved for \(h\) and \(k\).
- By multiplying the first equation by \(3\) and subtracting the second, \(10h = -10\) results in \(h = -1\).
- Substituting \(h = -1\) into \(3h - k = 1\) gives \(k = -4\), establishing the values of \(h\) and \(k\).
- The equations of the two lines given are \(5x + y - 4 = 0\) and \(3x + 4y - 4 = 0\), which define the relationship between them.
- To find the equation of a line segment between two points, use the midpoint formula, where the midpoint is \((\frac{\alpha_1 + \alpha_2}{2}, \frac{b_1 + b_2}{2})\).
- The equations \(5\alpha_1 + 4b_1 = 4\) and \(3\alpha_2 + 4b_2 - 4 = 0\) are used to derive the coordinates of the intersection points.
- The final equation is derived from substituting values back into the original equations, leading to a simplified form that can be solved for the desired variables.
