Straight Lines — Class 11 Maths Notes

Straight Lines · Class 11 Maths · 12 topics.

These notes are free to read without an account. Work through them in order, or use the chapter list to revise selectively before a test.

Topics covered in Straight Lines

  1. 1.Introduction of Straight Lines

    A straight line is one of the most fundamental concepts in geometry. In the Cartesian coordinate system, a straight line can be defined using an equation. Understanding straight lines is essential because they form the basis for many other geometrical shapes and concepts.

    Equation of a Straight Line

    The general form of the equation of a straight line in the Cartesian plane is: 𝐴𝑥+𝐵𝑦+𝐶=0Ax+By+C=0

    Here:

    • 𝐴A, 𝐵B, and 𝐶C are constants.
    • 𝑥x and 𝑦y are variables representing coordinates of any point on the line.

    Slope-Intercept Form

    A common way to express the equation of a straight line is the slope-intercept form: 𝑦=𝑚𝑥+𝑐y=mx+c

    Where:

    • 𝑚m is the slope of the line, representing its steepness.
    • 𝑐c is the y-intercept, the point where the line crosses the y-axis.

    Finding the Slope

    The slope (𝑚m) of a line can be found using two points (𝑥1,𝑦1)(x1​,y1​) and (𝑥2,𝑦2)(x2​,y2​) on the line: 𝑚=𝑦2−𝑦1𝑥2−𝑥1m=x2​−x1​y2​−y1​​

    Real-Life Example

    Imagine you are on a road trip, and your car's journey is represented by a straight line on a map. The slope of this line indicates how steep the road is. If you know two points along your route (e.g., two towns you pass through), you can calculate the slope to understand the incline or decline of the road.

    Step-by-Step Derivation of Slope-Intercept Form

    1. Start with two points on the line: (𝑥1,𝑦1)(x1​,y1​) and (𝑥2,𝑦2)(x2​,y2​).

    2. Calculate the slope: 𝑚=𝑦2−𝑦1𝑥2−𝑥1m=x2​−x1​y2​−y1​​

    3. Use the point-slope form of the line equation: 𝑦−𝑦1=𝑚(𝑥−𝑥1)y−y1​=m(x−x1​)

    4. Rearrange to get the slope-intercept form: 𝑦=𝑚𝑥−𝑚𝑥1+𝑦1y=mx−mx1​+y1​ Here, −𝑚𝑥1+𝑦1−mx1​+y1​ is a constant, which we denote as 𝑐c.

      So, the equation becomes: 𝑦=𝑚𝑥+𝑐y=mx+c

    Activity

    1. Identify two points on a graph: Plot points (2,3)(2,3) and (4,7)(4,7).

    2. Calculate the slope: 𝑚=7−34−2=42=2m=4−27−3​=24​=2

    3. Use one point to find the y-intercept: Using (2,3)(2,3): 3=2(2)+𝑐3=2(2)+c 3=4+𝑐3=4+c 𝑐=−1c=−1

    4. Write the equation: 𝑦=2𝑥−1y=2x−1

    Applications in Careers

    1. Engineering: Engineers use the concept of straight lines in designing and constructing buildings, bridges, and other structures to ensure they are stable and aligned correctly.

    2. Economics: Economists use straight-line graphs to represent supply and demand curves, showing relationships between quantities and prices.

    3. Computer Graphics: In computer graphics, straight lines are fundamental in rendering shapes and creating digital images.

    Summary

    Understanding straight lines involves learning about their equations, calculating slopes, and interpreting their graphs. This knowledge is essential in various fields, including engineering, economics, and computer graphics.

  2. 2.Important Formulae for Straight Lines

    Formulae for Straight Lines

    1. Slope of a Line: 𝑚=𝑦2−𝑦1𝑥2−𝑥1m=x2​−x1​y2​−y1​​

    2. Slope-Intercept Form: 𝑦=𝑚𝑥+𝑐y=mx+c

    3. Point-Slope Form: 𝑦−𝑦1=𝑚(𝑥−𝑥1)y−y1​=m(x−x1​)

    4. Two-Point Form: 𝑦−𝑦1=𝑦2−𝑦1𝑥2−𝑥1(𝑥−𝑥1)y−y1​=x2​−x1​y2​−y1​​(x−x1​)

    5. Intercept Form: 𝑥𝑎+𝑦𝑏=1ax​+by​=1

    6. General Form: 𝐴𝑥+𝐵𝑦+𝐶=0Ax+By+C=0

    7. Distance of a Point from a Line: 𝑑=∣𝐴𝑥1+𝐵𝑦1+𝐶∣𝐴2+𝐵2d=A2+B2​∣Ax1​+By1​+C∣​

    8. Angle Between Two Lines: tan⁡𝜃=∣𝑚1−𝑚21+𝑚1𝑚2∣tanθ=∣∣​1+m1​m2​m1​−m2​​∣∣​

    9. Condition for Parallel Lines: 𝑚1=𝑚2m1​=m2​

    10. Condition for Perpendicular Lines: 𝑚1⋅𝑚2=−1m1​⋅m2​=−1

    Detailed Explanation with Examples

    1. Slope of a Line

    The slope of a line passing through two points (𝑥1,𝑦1)(x1​,y1​) and (𝑥2,𝑦2)(x2​,y2​) is given by: 𝑚=𝑦2−𝑦1𝑥2−𝑥1m=x2​−x1​y2​−y1​​

    Example: Find the slope of the line passing through points (3, 4) and (7, 8). 𝑚=8−47−3=44=1m=7−38−4​=44​=1

    Application: The slope helps in determining the steepness or incline of roads and railways.

    2. Slope-Intercept Form

    The equation of a line in slope-intercept form is: 𝑦=𝑚𝑥+𝑐y=mx+c where 𝑚m is the slope and 𝑐c is the y-intercept.

    Example: A line has a slope of 2 and passes through the point (0, -3). Its equation is: 𝑦=2𝑥−3y=2x−3

    Application: This form is commonly used in economics to represent cost functions and in physics to describe linear motion.

    3. Point-Slope Form

    The equation of a line with slope 𝑚m passing through a point (𝑥1,𝑦1)(x1​,y1​) is: 𝑦−𝑦1=𝑚(𝑥−𝑥1)y−y1​=m(x−x1​)

    Example: Find the equation of a line with slope 3 passing through point (1, 2). 𝑦−2=3(𝑥−1)y−2=3(x−1) 𝑦=3𝑥−3+2y=3x−3+2 𝑦=3𝑥−1y=3x−1

    Application: Used in situations where a specific point on the line is known, such as in navigation and mapping.

    4. Two-Point Form

    The equation of a line passing through two points (𝑥1,𝑦1)(x1​,y1​) and (𝑥2,𝑦2)(x2​,y2​) is: 𝑦−𝑦1=𝑦2−𝑦1𝑥2−𝑥1(𝑥−𝑥1)y−y1​=x2​−x1​y2​−y1​​(x−x1​)

    Example: Find the equation of the line passing through points (1, 2) and (3, 6). 𝑦−2=6−23−1(𝑥−1)y−2=3−16−2​(x−1) 𝑦−2=2(𝑥−1)y−2=2(x−1) 𝑦=2𝑥−2+2y=2x−2+2 𝑦=2𝑥y=2x

    Application: Commonly used in computer graphics and geometry.

    5. Intercept Form

    The equation of a line that intercepts the x-axis at 𝑎a and the y-axis at 𝑏b is: 𝑥𝑎+𝑦𝑏=1ax​+by​=1

    Example: Find the equation of a line that intercepts the x-axis at 4 and the y-axis at 3. 𝑥4+𝑦3=14x​+3y​=1

    Application: Used in graphing linear inequalities and budget constraints.

    6. General Form

    The general form of the equation of a line is: 𝐴𝑥+𝐵𝑦+𝐶=0Ax+By+C=0

    Example: Convert the slope-intercept form 𝑦=2𝑥−3y=2x−3 to the general form. 2𝑥−𝑦−3=02x−y−3=0

    Application: This form is useful in algebraic manipulations and in systems of linear equations.

    7. Distance of a Point from a Line

    The distance of a point (𝑥1,𝑦1)(x1​,y1​) from a line 𝐴𝑥+𝐵𝑦+𝐶=0Ax+By+C=0 is: 𝑑=∣𝐴𝑥1+𝐵𝑦1+𝐶∣𝐴2+𝐵2d=A2+B2​∣Ax1​+By1​+C∣​

    Example: Find the distance of the point (3, 4) from the line 3𝑥+4𝑦−5=03x+4y−5=0. 𝑑=∣3(3)+4(4)−5∣32+42d=32+42​∣3(3)+4(4)−5∣​ 𝑑=∣9+16−5∣9+16d=9+16​∣9+16−5∣​ 𝑑=205=4d=520​=4

    Application: This formula is used in construction and navigation to find the shortest path.

    8. Angle Between Two Lines

    The angle 𝜃θ between two lines with slopes 𝑚1m1​ and 𝑚2m2​ is: tan⁡𝜃=∣𝑚1−𝑚21+𝑚1𝑚2∣tanθ=∣∣​1+m1​m2​m1​−m2​​∣∣​

    Example: Find the angle between the lines with slopes 1 and -1. tan⁡𝜃=∣1−(−1)1+1(−1)∣=∣20∣tanθ=∣∣​1+1(−1)1−(−1)​∣∣​=∣∣​02​∣∣​ The lines are perpendicular, so 𝜃=90∘θ=90∘.

    Application: Used in road design and architecture to determine angles between paths or walls.

    9. Condition for Parallel Lines

    Two lines are parallel if their slopes are equal: 𝑚1=𝑚2m1​=m2​

    Example: If the slope of one line is 2, the slope of a parallel line is also 2.

    Application: Parallel lines are used in construction to ensure walls and beams are aligned properly.

    10. Condition for Perpendicular Lines

    Two lines are perpendicular if the product of their slopes is -1: 𝑚1⋅𝑚2=−1m1​⋅m2​=−1

    Example: If the slope of one line is 3, the slope of a perpendicular line is: 𝑚2=−13m2​=−31​

    Application: Perpendicular lines are essential in creating right angles in various designs and constructions.

  3. 3.Slope of a Line and Its Definition

    Definition

    The slope of a line is a measure of its steepness and direction. It is defined as the ratio of the vertical change (rise) to the horizontal change (run) between any two points on the line. Mathematically, if you have two points (𝑥1,𝑦1)(x1​,y1​) and (𝑥2,𝑦2)(x2​,y2​) on a line, the slope 𝑚m is given by:

    𝑚=𝑦2−𝑦1𝑥2−𝑥1m=x2​−x1​y2​−y1​​

    The slope tells us how much 𝑦y changes for a unit change in 𝑥x.

    Slope of a Line When Coordinates of Any Two Points on the Line Are Given

    Given two points (𝑥1,𝑦1)(x1​,y1​) and (𝑥2,𝑦2)(x2​,y2​) on a line, the slope 𝑚m can be calculated using the formula:

    𝑚=𝑦2−𝑦1𝑥2−𝑥1m=x2​−x1​y2​−y1​​

    Example

    Let's find the slope of a line passing through the points (3, 4) and (7, 8):

    1. Identify the coordinates: (𝑥1,𝑦1)=(3,4)(x1​,y1​)=(3,4) (𝑥2,𝑦2)=(7,8)(x2​,y2​)=(7,8)

    2. Substitute these values into the slope formula: 𝑚=8−47−3=44=1m=7−38−4​=44​=1

    So, the slope of the line is 1.

    Conditions for Parallelism and Perpendicularity of Lines in Terms of Their Slopes

    Parallel Lines

    Two lines are parallel if they have the same slope. This means their direction is the same, and they will never intersect.

    If the slopes of two lines are 𝑚1m1​ and 𝑚2m2​, then the lines are parallel if: 𝑚1=𝑚2m1​=m2​

    Example

    If the slope of one line is 2, the slope of a line parallel to it must also be 2.

    Perpendicular Lines

    Two lines are perpendicular if the product of their slopes is -1. This means they intersect at a right angle (90 degrees).

    If the slopes of two lines are 𝑚1m1​ and 𝑚2m2​, then the lines are perpendicular if: 𝑚1⋅𝑚2=−1m1​⋅m2​=−1

    Example

    If the slope of one line is 3, the slope of a line perpendicular to it must be: 𝑚2=−13m2​=−31​

    Angle Between Two Lines

    The angle 𝜃θ between two lines with slopes 𝑚1m1​ and 𝑚2m2​ can be found using the following formula: tan⁡𝜃=∣𝑚1−𝑚21+𝑚1𝑚2∣tanθ=∣∣​1+m1​m2​m1​−m2​​∣∣​

    Example

    Let's find the angle between two lines with slopes 𝑚1=1m1​=1 and 𝑚2=−1m2​=−1:

    1. Substitute the slopes into the formula: tan⁡𝜃=∣1−(−1)1+1(−1)∣=∣1+11−1∣=∣20∣tanθ=∣∣​1+1(−1)1−(−1)​∣∣​=∣∣​1−11+1​∣∣​=∣∣​02​∣∣​

    2. Since the denominator is 0, the angle 𝜃θ is 90 degrees.

    In this case, the lines are perpendicular, as expected from their slope condition.

    Real-Life Example

    Imagine you are an architect designing a staircase. The slope of the stairs (also called the pitch) needs to be calculated to ensure it meets building regulations. If you know two points on the edge of the staircase, you can calculate the slope to ensure it's not too steep.

    Summary

    Understanding the slope of a line is crucial for various mathematical and real-life applications. It helps in determining the steepness and direction of lines, analyzing the conditions for parallelism and perpendicularity, and calculating the angle between two lines.

  4. 4.Exercise Questions

    Question 1

    Draw a quadrilateral in the Cartesian plane, whose vertices are (−4,5)(−4,5), (0,7)(0,7), (5,−5)(5,−5), and (−4,−2)(−4,−2). Also, find its area.

    Step-by-Step Solution

    1. Plotting the vertices on the Cartesian plane:

      • (−4,5)(−4,5)
      • (0,7)(0,7)
      • (5,−5)(5,−5)
      • (−4,−2)(−4,−2)

    2. Connecting the vertices to form the quadrilateral:

    3. Finding the area using the Shoelace formula (or Gauss's area formula for polygons): Area=12∣𝑥1𝑦2+𝑥2𝑦3+𝑥3𝑦4+𝑥4𝑦1−(𝑦1𝑥2+𝑦2𝑥3+𝑦3𝑥4+𝑦4𝑥1)∣Area=21​∣x1​y2​+x2​y3​+x3​y4​+x4​y1​−(y1​x2​+y2​x3​+y3​x4​+y4​x1​)∣

      Substituting the given points: (𝑥1,𝑦1)=(−4,5)(x1​,y1​)=(−4,5) (𝑥2,𝑦2)=(0,7)(x2​,y2​)=(0,7) (𝑥3,𝑦3)=(5,−5)(x3​,y3​)=(5,−5) (𝑥4,𝑦4)=(−4,−2)(x4​,y4​)=(−4,−2)

      Calculate: Area=12∣(−4⋅7)+(0⋅−5)+(5⋅−2)+(−4⋅5)−(5⋅0)−(7⋅5)−(−5⋅−4)−(−2⋅−4)∣Area=21​∣(−4⋅7)+(0⋅−5)+(5⋅−2)+(−4⋅5)−(5⋅0)−(7⋅5)−(−5⋅−4)−(−2⋅−4)∣ =12∣(−28)+0+(−10)+(−20)−0−35−20−8∣=21​∣(−28)+0+(−10)+(−20)−0−35−20−8∣ =12∣−28−10−20−35−20−8∣=21​∣−28−10−20−35−20−8∣ =12∣−121∣=21​∣−121∣ =12⋅121=21​⋅121 =60.5=60.5

      So, the area of the quadrilateral is 60.5 square units.

    ------------------------------------------------------------------------------------------------------------------------------------------ Question 2

    The base of an equilateral triangle with side 2𝑎2a lies along the y-axis such that the midpoint of the base is at the origin. Find the vertices of the triangle.

    Step-by-Step Solution

    1. Base of the equilateral triangle:

      • The side of the base is 2𝑎2a, and the midpoint of the base is at the origin (0,0)(0,0).
      • The coordinates of the endpoints of the base will be (0,𝑎)(0,a) and (0,−𝑎)(0,−a) because they are equidistant from the origin along the y-axis.

    2. Finding the third vertex:

      • Since it's an equilateral triangle, the distance between any two vertices is 2𝑎2a.
      • The coordinates of the third vertex (𝑥,𝑦)(x,y) must satisfy the condition of being 2𝑎2a units away from both (0,𝑎)(0,a) and (0,−𝑎)(0,−a).

      Using the distance formula: 𝑥2+(𝑦−𝑎)2=2𝑎x2+(y−a)2​=2a 𝑥2+(𝑦+𝑎)2=2𝑎x2+(y+a)2​=2a

      Solving these equations: 𝑥2+(𝑦−𝑎)2=4𝑎2x2+(y−a)2=4a2 𝑥2+(𝑦+𝑎)2=4𝑎2x2+(y+a)2=4a2

      Expanding and simplifying: 𝑥2+𝑦2−2𝑎𝑦+𝑎2=4𝑎2x2+y2−2ay+a2=4a2 𝑥2+𝑦2+2𝑎𝑦+𝑎2=4𝑎2x2+y2+2ay+a2=4a2

      Subtracting the first equation from the second: 4𝑎𝑦=04ay=0 𝑦=0y=0

      Since 𝑦y cannot be zero (it would mean the third vertex lies on the x-axis, which is not possible), solving for 𝑥x and 𝑦y: 𝑥2+𝑦2=3𝑎2x2+y2=3a2

      This means 𝑦y can be positive or negative. Hence, 𝑦=3𝑎y=3​a

      The coordinates of the third vertex are (±𝑎,3𝑎)(±a,3​a).

      Therefore, the vertices of the equilateral triangle are:

      • (0,𝑎)(0,a)
      • (0,−𝑎)(0,−a)
      • (±3𝑎,0)(±3​a,0)

  5. 5.Exercise Questions

    Question 1

    Find a point on the x-axis, which is equidistant from the points (7,6)(7,6) and (3,4)(3,4).

    Solution

    A point on the x-axis has coordinates (𝑥,0)(x,0). This point is equidistant from (7,6)(7,6) and (3,4)(3,4). This means the distance from (𝑥,0)(x,0) to (7,6)(7,6) is equal to the distance from (𝑥,0)(x,0) to (3,4)(3,4).

    Using the distance formula: (𝑥−7)2+(0−6)2=(𝑥−3)2+(0−4)2(x−7)2+(0−6)2​=(x−3)2+(0−4)2​

    Squaring both sides to remove the square roots: (𝑥−7)2+62=(𝑥−3)2+42(x−7)2+62=(x−3)2+42

    Simplify and solve for 𝑥x: (𝑥−7)2+36=(𝑥−3)2+16(x−7)2+36=(x−3)2+16 𝑥2−14𝑥+49+36=𝑥2−6𝑥+9+16x2−14x+49+36=x2−6x+9+16 𝑥2−14𝑥+85=𝑥2−6𝑥+25x2−14x+85=x2−6x+25

    Cancel 𝑥2x2 from both sides: −14𝑥+85=−6𝑥+25−14x+85=−6x+25 85−25=14𝑥−6𝑥85−25=14x−6x 60=8𝑥60=8x 𝑥=608x=860​ 𝑥=7.5x=7.5

    So, the point on the x-axis that is equidistant from (7,6)(7,6) and (3,4)(3,4) is (7.5,0)(7.5,0).

    Question 2

    Find the slope of a line, which passes through the origin, and the mid-point of the line segment joining the points 𝑃(0,−4)P(0,−4) and 𝐵(8,0)B(8,0).

    Solution

    1. Find the mid-point of the line segment joining points 𝑃(0,−4)P(0,−4) and 𝐵(8,0)B(8,0): The mid-point formula is: (𝑥1+𝑥22,𝑦1+𝑦22)(2x1​+x2​​,2y1​+y2​​)

      Substituting the given points: (0+82,−4+02)=(4,−2)(20+8​,2−4+0​)=(4,−2)

    2. Find the slope of the line passing through the origin (0,0)(0,0) and the mid-point (4,−2)(4,−2): The slope 𝑚m is given by: 𝑚=𝑦2−𝑦1𝑥2−𝑥1m=x2​−x1​y2​−y1​​

      Substituting the points: 𝑚=−2−04−0=−24=−12m=4−0−2−0​=4−2​=−21​

    So, the slope of the line is −12−21​.

    Question 3

    Without using the Pythagoras theorem, show that the points (4,4)(4,4), (3,5)(3,5), and (−1,−1)(−1,−1) are the vertices of a right-angled triangle.

    Solution

    To show that the points form a right-angled triangle, we need to show that the slopes of two sides are negative reciprocals of each other (i.e., the product of their slopes is -1), indicating a right angle.

    1. Find the slopes of the sides:

      • Slope of the line segment joining (4,4)(4,4) and (3,5)(3,5): 𝑚1=5−43−4=1−1=−1m1​=3−45−4​=−11​=−1

      • Slope of the line segment joining (3,5)(3,5) and (−1,−1)(−1,−1): 𝑚2=−1−5−1−3=−6−4=32m2​=−1−3−1−5​=−4−6​=23​

      • Slope of the line segment joining (4,4)(4,4) and (−1,−1)(−1,−1): 𝑚3=−1−4−1−4=−5−5=1m3​=−1−4−1−4​=−5−5​=1

    2. Check for perpendicularity:

      • The product of slopes 𝑚1m1​ and 𝑚3m3​: 𝑚1⋅𝑚3=−1⋅1=−1m1​⋅m3​=−1⋅1=−1

    Since the product of slopes 𝑚1m1​ and 𝑚3m3​ is -1, the line segments joining (4,4)(4,4) and (3,5)(3,5), and (4,4)(4,4) and (−1,−1)(−1,−1) are perpendicular, indicating a right angle at point (4,4)(4,4).

    Thus, the points (4,4)(4,4), (3,5)(3,5), and (−1,−1)(−1,−1) are the vertices of a right-angled triangle.

  6. 6.Exercise Questions

    Question 01

    Without using distance formula, show that points (−2,−1)(−2,−1), (4,0)(4,0), (3,3)(3,3), and (−3,2)(−3,2) are the vertices of a parallelogram.

    Solution

    To show that the given points are vertices of a parallelogram, we need to show that the midpoints of the diagonals are the same.

    1. Find the midpoints of the diagonals:

      • Diagonal 1: Joining points (−2,−1)(−2,−1) and (3,3)(3,3) Midpoint=(−2+32,−1+32)=(12,1)Midpoint=(2−2+3​,2−1+3​)=(21​,1)

      • Diagonal 2: Joining points (4,0)(4,0) and (−3,2)(−3,2) Midpoint=(4−32,0+22)=(12,1)Midpoint=(24−3​,20+2​)=(21​,1)

    Since the midpoints of the diagonals are the same, the given points are the vertices of a parallelogram.

    Question 02

    Find the angle between the x-axis and the line joining the points (3,−1)(3,−1) and (4,−2)(4,−2).

    Solution

    1. Find the slope of the line joining the points (3,−1)(3,−1) and (4,−2)(4,−2): 𝑚=−2−(−1)4−3=−2+11=−11=−1m=4−3−2−(−1)​=1−2+1​=1−1​=−1

    2. Find the angle 𝜃θ between the line and the x-axis: The slope 𝑚m is the tangent of the angle 𝜃θ between the line and the x-axis. tan⁡𝜃=𝑚=−1tanθ=m=−1 𝜃=tan⁡−1(−1)θ=tan−1(−1) 𝜃=−45∘ or 135∘θ=−45∘ or 135∘

    So, the angle between the x-axis and the line is either −45∘−45∘ or 135∘135∘.

    Question 03

    The slope of a line is double of the slope of another line. If the tangent of the angle between them is 1331​, find the slopes of the lines.

    Solution

    Let the slopes of the two lines be 𝑚1m1​ and 𝑚2m2​, with 𝑚1=2𝑚2m1​=2m2​.

    The tangent of the angle 𝜃θ between the two lines is given by: tan⁡𝜃=∣𝑚1−𝑚21+𝑚1𝑚2∣=13tanθ=∣∣​1+m1​m2​m1​−m2​​∣∣​=31​

    Substitute 𝑚1=2𝑚2m1​=2m2​: ∣2𝑚2−𝑚21+2𝑚2⋅𝑚2∣=13∣∣​1+2m2​⋅m2​2m2​−m2​​∣∣​=31​ ∣𝑚21+2𝑚22∣=13∣∣​1+2m22​m2​​∣∣​=31​

    Solve for 𝑚2m2​: 𝑚21+2𝑚22=13 or −𝑚21+2𝑚22=131+2m22​m2​​=31​ or 1+2m22​−m2​​=31​

    For the positive case: 3𝑚2=1+2𝑚223m2​=1+2m22​ 2𝑚22−3𝑚2+1=02m22​−3m2​+1=0

    Solve the quadratic equation: 𝑚2=3±9−84m2​=43±9−8​​ 𝑚2=3±14m2​=43±1​ 𝑚2=1 or 𝑚2=12m2​=1 or m2​=21​

    For the negative case: −3𝑚2=1+2𝑚22−3m2​=1+2m22​ 2𝑚22+3𝑚2+1=02m22​+3m2​+1=0

    Solve the quadratic equation: 𝑚2=−3±9−84m2​=4−3±9−8​​ 𝑚2=−3±14m2​=4−3±1​ 𝑚2=−1 or 𝑚2=−12m2​=−1 or m2​=−21​

    So, the possible slopes are 𝑚2=1m2​=1 or 𝑚2=12m2​=21​, and 𝑚1=2m1​=2 or 𝑚1=1m1​=1.

    Question 04

    A line passes through (𝑥1,𝑦1)(x1​,y1​) and (ℎ,𝑘)(h,k). If slope of the line is 𝑚m, show that 𝑘−𝑦1=𝑚(ℎ−𝑥1)k−y1​=m(h−x1​).

    Solution

    The slope 𝑚m of the line passing through the points (𝑥1,𝑦1)(x1​,y1​) and (ℎ,𝑘)(h,k) is given by: 𝑚=𝑘−𝑦1ℎ−𝑥1m=h−x1​k−y1​​

    Rearrange the equation to show that 𝑘−𝑦1=𝑚(ℎ−𝑥1)k−y1​=m(h−x1​): 𝑚(ℎ−𝑥1)=𝑘−𝑦1m(h−x1​)=k−y1​ 𝑘−𝑦1=𝑚(ℎ−𝑥1)k−y1​=m(h−x1​)

    Hence, it is shown that 𝑘−𝑦1=𝑚(ℎ−𝑥1)k−y1​=m(h−x1​).

  7. 7.Various Forms of the Equation of a Line

    Understanding different forms of the equation of a line is essential in mathematics. Each form is used depending on the information available and the application.

    1. Horizontal and Vertical Lines

    Horizontal Line

    A horizontal line has a constant 𝑦y-coordinate for all points on the line. The equation of a horizontal line passing through the point (𝑥,𝑦)(x,y) is:

    𝑦=𝑐y=c

    where 𝑐c is the constant 𝑦y-coordinate.

    Example

    The equation of a horizontal line passing through (3,4)(3,4) is: 𝑦=4y=4

    Vertical Line

    A vertical line has a constant 𝑥x-coordinate for all points on the line. The equation of a vertical line passing through the point (𝑥,𝑦)(x,y) is:

    𝑥=𝑐x=c

    where 𝑐c is the constant 𝑥x-coordinate.

    Example

    The equation of a vertical line passing through (3,4)(3,4) is: 𝑥=3x=3

    2. Point-Slope Form

    The point-slope form is useful when you know the slope of the line and one point on the line. The equation is:

    𝑦−𝑦1=𝑚(𝑥−𝑥1)y−y1​=m(x−x1​)

    where:

    • 𝑚m is the slope of the line.
    • (𝑥1,𝑦1)(x1​,y1​) is a point on the line.
    Example

    For a line with a slope of 2 passing through the point (3,4)(3,4): 𝑦−4=2(𝑥−3)y−4=2(x−3)

    3. Two-Point Form

    The two-point form is used when you know two points on the line. The equation is:

    𝑦−𝑦1=𝑦2−𝑦1𝑥2−𝑥1(𝑥−𝑥1)y−y1​=x2​−x1​y2​−y1​​(x−x1​)

    where (𝑥1,𝑦1)(x1​,y1​) and (𝑥2,𝑦2)(x2​,y2​) are two points on the line.

    Example

    For a line passing through the points (1,2)(1,2) and (4,6)(4,6): 𝑦−2=6−24−1(𝑥−1)y−2=4−16−2​(x−1) 𝑦−2=43(𝑥−1)y−2=34​(x−1)

    4. Slope-Intercept Form

    The slope-intercept form is one of the most commonly used forms. It is useful when you know the slope of the line and the y-intercept. The equation is:

    𝑦=𝑚𝑥+𝑐y=mx+c

    where:

    • 𝑚m is the slope of the line.
    • 𝑐c is the y-intercept (the point where the line crosses the y-axis).
    Example

    For a line with a slope of 3 and a y-intercept of -2: 𝑦=3𝑥−2y=3x−2

    5. Intercept Form

    The intercept form is used when you know the x-intercept and the y-intercept of the line. The equation is:

    𝑥𝑎+𝑦𝑏=1ax​+by​=1

    where:

    • 𝑎a is the x-intercept.
    • 𝑏b is the y-intercept.
    Example

    For a line with an x-intercept of 4 and a y-intercept of 3: 𝑥4+𝑦3=14x​+3y​=1

  8. 8.Distance of a Point from a Line

    To find the perpendicular distance of a point

    𝑃(𝑥1,𝑦1)P(x1​,y1​) from a line given by the equation 𝐴𝑥+𝐵𝑦+𝐶=0Ax+By+C=0, we use the formula:

    𝑑=∣𝐴𝑥1+𝐵𝑦1+𝐶∣𝐴2+𝐵2d=A2+B2​∣Ax1​+By1​+C∣​

    Let's break down the steps to understand how this formula is derived and how to use it with an example. The given image will help us visualize the process.

    Step-by-Step Explanation

    1. Equation of the Line: The line 𝐿L is given by the equation 𝐴𝑥+𝐵𝑦+𝐶=0Ax+By+C=0.

    2. Coordinates of the Point: We need to find the distance from the point 𝑃(𝑥1,𝑦1)P(x1​,y1​) to the line 𝐿L.

    3. Perpendicular Distance Formula: The formula to find the perpendicular distance 𝑑d of the point 𝑃P from the line 𝐿L is: 𝑑=∣𝐴𝑥1+𝐵𝑦1+𝐶∣𝐴2+𝐵2d=A2+B2​∣Ax1​+By1​+C∣​

      This formula comes from the geometric property that the shortest distance from a point to a line is the length of the perpendicular dropped from the point to the line.

    Example

    Let's find the distance from the point 𝑃(3,4)P(3,4) to the line given by the equation 3𝑥+4𝑦−5=03x+4y−5=0.

    1. Identify the coefficients and coordinates:

      • 𝐴=3A=3
      • 𝐵=4B=4
      • 𝐶=−5C=−5
      • 𝑥1=3x1​=3
      • 𝑦1=4y1​=4

    2. Substitute these values into the distance formula: 𝑑=∣3(3)+4(4)−5∣32+42d=32+42​∣3(3)+4(4)−5∣​ 𝑑=∣9+16−5∣9+16d=9+16​∣9+16−5∣​ 𝑑=∣20∣25d=25​∣20∣​ 𝑑=205d=520​ 𝑑=4d=4

    So, the perpendicular distance from the point 𝑃(3,4)P(3,4) to the line 3𝑥+4𝑦−5=03x+4y−5=0 is 4 units.

    Geometric Interpretation Using the Image

    1. Line Equation: The line 𝐿L is represented by the equation 𝐴𝑥+𝐵𝑦+𝐶=0Ax+By+C=0.

    2. Points of Interest:

      • Point 𝑃(𝑥1,𝑦1)P(x1​,y1​) is the point from which we want to find the distance to the line.
      • Point 𝑅(0,−𝐶𝐵)R(0,−BC​) is where the line intersects the y-axis (y-intercept).
      • Point 𝑄(−𝐶𝐴,0)Q(−AC​,0) is where the line intersects the x-axis (x-intercept).

    3. Perpendicular Distance: The perpendicular distance 𝑑d is the shortest distance from point 𝑃P to the line 𝐿L, shown as the segment 𝑃𝑀PM in the diagram, where 𝑀M is the foot of the perpendicular from 𝑃P to 𝐿L.

    4. Visualization:

      • The point 𝑀M is the projection of 𝑃P on the line 𝐿L, making 𝑃𝑀PM perpendicular to 𝐿L.
      • This perpendicular distance can be calculated using the given formula, ensuring the absolute value to handle distances correctly (negative signs for direction do not affect distance).

    By understanding and visualizing these concepts with the help of the formula and the diagram, you can easily calculate the distance of any point from a line in a Cartesian plane.

  9. 9.Distance Between Two Parallel Lines

    To find the distance between two parallel lines, let's consider the general forms of the two lines. Parallel lines have the same slope but different y-intercepts. The equations of the two parallel lines can be written as:

    𝑦=𝑚𝑥+𝑐1y=mx+c1​ 𝑦=𝑚𝑥+𝑐2y=mx+c2​

    Here, 𝑚m is the slope of both lines, and 𝑐1c1​ and 𝑐2c2​ are the y-intercepts.

    The distance 𝑑d between these two parallel lines can be found using the following formula:

    𝑑=∣𝑐2−𝑐1∣1+𝑚2d=1+m2​∣c2​−c1​∣​

    Let's break down the steps to understand how this formula is derived and how to use it with an example. The given image will help us visualize the process.

    Derivation and Explanation

    1. Equation of the Lines: The lines are given by: 𝑦=𝑚𝑥+𝑐1y=mx+c1​ 𝑦=𝑚𝑥+𝑐2y=mx+c2​

    2. Distance Formula: The perpendicular distance 𝑑d between two parallel lines 𝑦=𝑚𝑥+𝑐1y=mx+c1​ and 𝑦=𝑚𝑥+𝑐2y=mx+c2​ is given by: 𝑑=∣𝑐2−𝑐1∣1+𝑚2d=1+m2​∣c2​−c1​∣​

    Example

    Let's find the distance between the lines 𝑦=2𝑥+3y=2x+3 and 𝑦=2𝑥−1y=2x−1.

    1. Identify the coefficients and intercepts:

      • 𝑚=2m=2
      • 𝑐1=3c1​=3
      • 𝑐2=−1c2​=−1

    2. Substitute these values into the distance formula: 𝑑=∣𝑐2−𝑐1∣1+𝑚2d=1+m2​∣c2​−c1​∣​ 𝑑=∣−1−3∣1+22d=1+22​∣−1−3∣​ 𝑑=∣−4∣1+4d=1+4​∣−4∣​ 𝑑=45d=5​4​ 𝑑=45⋅55d=5​4​⋅5​5​​ 𝑑=455d=545​​

    So, the distance between the lines 𝑦=2𝑥+3y=2x+3 and 𝑦=2𝑥−1y=2x−1 is 455545​​ units.

    Visual Explanation Using the Image

    1. Lines Representation:

      • The two parallel lines are represented as 𝑦=𝑚𝑥+𝑐1y=mx+c1​ and 𝑦=𝑚𝑥+𝑐2y=mx+c2​.
      • The lines have the same slope 𝑚m but different y-intercepts 𝑐1c1​ and 𝑐2c2​.

    2. Point on One Line:

      • Let's take a point 𝐴(−𝑐1𝑚,0)A(−mc1​​,0) on the line 𝑦=𝑚𝑥+𝑐1y=mx+c1​.

    3. Distance 𝑑d:

      • The perpendicular distance from this point 𝐴A to the line 𝑦=𝑚𝑥+𝑐2y=mx+c2​ is the shortest distance between the two parallel lines.
      • This distance 𝑑d can be computed using the given formula, considering the absolute difference in intercepts and the slope.

  10. 10.Exercise Questions

    Question 1

    Reduce the following equations into slope-intercept form and find their slopes and the y-intercepts.

    (i) 𝑥+7𝑦=0x+7y=0

    (ii) 6𝑥+3𝑦−5=06x+3y−5=0

    (iii) 𝑦=0y=0

    Solution

    (i) Equation: 𝑥+7𝑦=0x+7y=0

    1. Rewrite in slope-intercept form 𝑦=𝑚𝑥+𝑐y=mx+c: 7𝑦=−𝑥7y=−x 𝑦=−17𝑥y=−71​x

    2. Slope (m): −17−71​

    3. y-intercept (c): 0

    (ii) Equation: 6𝑥+3𝑦−5=06x+3y−5=0

    1. Rewrite in slope-intercept form 𝑦=𝑚𝑥+𝑐y=mx+c: 3𝑦=−6𝑥+53y=−6x+5 𝑦=−2𝑥+53y=−2x+35​

    2. Slope (m): −2−2

    3. y-intercept (c): 5335​

    (iii) Equation: 𝑦=0y=0

    1. This is already in slope-intercept form 𝑦=𝑚𝑥+𝑐y=mx+c: 𝑦=0y=0

    2. Slope (m): 0

    3. y-intercept (c): 0

    Question 2

    Reduce the following equations into intercept form and find their intercepts on the axes.

    (i) 3𝑥+2𝑦−12=03x+2y−12=0

    (ii) 4𝑥−3𝑦=64x−3y=6

    (iii) 3𝑦+2=03y+2=0

    Solution

    (i) Equation: 3𝑥+2𝑦−12=03x+2y−12=0

    1. Rewrite in intercept form 𝑥𝑎+𝑦𝑏=1ax​+by​=1: 3𝑥+2𝑦=123x+2y=12 3𝑥12+2𝑦12=1123x​+122y​=1 𝑥4+𝑦6=14x​+6y​=1

    2. x-intercept (a): 4

    3. y-intercept (b): 6

    (ii) Equation: 4𝑥−3𝑦=64x−3y=6

    1. Rewrite in intercept form 𝑥𝑎+𝑦𝑏=1ax​+by​=1: 4𝑥6−3𝑦6=164x​−63y​=1 𝑥1.5−𝑦2=11.5x​−2y​=1 𝑥1.5+𝑦−2=11.5x​+−2y​=1

    2. x-intercept (a): 1.5

    3. y-intercept (b): -2

    (iii) Equation: 3𝑦+2=03y+2=0

    1. Rewrite in intercept form 𝑥𝑎+𝑦𝑏=1ax​+by​=1: 3𝑦=−23y=−2 𝑦=−23y=−32​

    2. The line is horizontal, so:

      • x-intercept (a): Undefined (no x-intercept)
      • y-intercept (b): −23−32​

    Question 3

    Find the distance of the point (−1,1)(−1,1) from the line 12𝑥+6𝑦=5(𝑦−2)12x+6y=5(y−2).

    Solution

    1. First, simplify the given line equation: 12𝑥+6𝑦=5𝑦−1012x+6y=5y−10 12𝑥+6𝑦−5𝑦=−1012x+6y−5y=−10 12𝑥+𝑦=−1012x+y=−10 12𝑥+𝑦+10=012x+y+10=0

    2. Use the distance formula 𝑑=∣𝐴𝑥1+𝐵𝑦1+𝐶∣𝐴2+𝐵2d=A2+B2​∣Ax1​+By1​+C∣​: 𝐴=12,𝐵=1,𝐶=10A=12,B=1,C=10 𝑥1=−1,𝑦1=1x1​=−1,y1​=1

    3. Substitute these values into the distance formula: 𝑑=∣12(−1)+1(1)+10∣122+12d=122+12​∣12(−1)+1(1)+10∣​ 𝑑=∣−12+1+10∣144+1d=144+1​∣−12+1+10∣​ 𝑑=∣−1∣145d=145​∣−1∣​ 𝑑=1145d=145​1​ 𝑑=1145⋅145145d=145​1​⋅145​145​​ 𝑑=145145d=145145​​

    So, the distance of the point (−1,1)(−1,1) from the line 12𝑥+𝑦+10=012x+y+10=0 is 145145145145​​ units.

  11. 11.Exercise Questions

    Question 01

    Find the angle between the lines 3𝑥+𝑦=13​x+y=1 and 𝑥+3𝑦=1x+3​y=1.

    Solution

    To find the angle between two lines, we first need to determine their slopes. The general form of a line is 𝐴𝑥+𝐵𝑦+𝐶=0Ax+By+C=0, and the slope 𝑚m of the line is given by −𝐴𝐵−BA​.

    1. Find the slopes of the lines:

      For the first line 3𝑥+𝑦=13​x+y=1: Slope (𝑚1)=−31=−3Slope (m1​)=−13​​=−3​

      For the second line 𝑥+3𝑦=1x+3​y=1: Slope (𝑚2)=−13=−13Slope (m2​)=−3​1​=−3​1​

    2. Use the formula for the angle 𝜃θ between two lines: tan⁡𝜃=∣𝑚1−𝑚21+𝑚1𝑚2∣tanθ=∣∣​1+m1​m2​m1​−m2​​∣∣​

      Substitute the slopes 𝑚1m1​ and 𝑚2m2​: tan⁡𝜃=∣−3−(−13)1+(−3)(−13)∣tanθ=∣∣​1+(−3​)(−3​1​)−3​−(−3​1​)​∣∣​ tan⁡𝜃=∣−3+131+1∣tanθ=∣∣​1+1−3​+3​1​​∣∣​ tan⁡𝜃=∣−3+132∣tanθ=∣∣​2−3​+3​1​​∣∣​ tan⁡𝜃=∣−3/3+1/32∣tanθ=∣∣​2−3/3​+1/3​​∣∣​ tan⁡𝜃=∣−2/32∣tanθ=∣∣​2−2/3​​∣∣​ tan⁡𝜃=∣−223∣tanθ=∣∣​23​−2​∣∣​ tan⁡𝜃=∣−13∣tanθ=∣∣​3​−1​∣∣​ tan⁡𝜃=13tanθ=3​1​

    3. Find the angle: 𝜃=tan⁡−1(13)θ=tan−1(3​1​) 𝜃=30∘θ=30∘

    So, the angle between the lines 3𝑥+𝑦=13​x+y=1 and 𝑥+3𝑦=1x+3​y=1 is 30∘30∘.

    Question 02

    The line through the points (ℎ,3)(h,3) and (4,1)(4,1) intersects the line 7𝑥−9𝑦−19=07x−9y−19=0 at a right angle. Find the value of ℎh.

    Solution

    To find the value of ℎh, we need to determine the slope of the line passing through (ℎ,3)(h,3) and (4,1)(4,1), and then use the condition for perpendicular lines.

    1. Find the slope of the line through (ℎ,3)(h,3) and (4,1)(4,1): Slope 𝑚1=1−34−ℎ=−24−ℎSlope m1​=4−h1−3​=4−h−2​

    2. Find the slope of the given line 7𝑥−9𝑦−19=07x−9y−19=0: Rewrite in slope-intercept form: 7𝑥−9𝑦=197x−9y=19 −9𝑦=−7𝑥+19−9y=−7x+19 𝑦=79𝑥−199y=97​x−919​ Slope 𝑚2=79Slope m2​=97​

    3. Use the condition for perpendicular lines: The product of the slopes of two perpendicular lines is −1−1: 𝑚1⋅𝑚2=−1m1​⋅m2​=−1 Substitute the slopes 𝑚1m1​ and 𝑚2m2​: (−24−ℎ)⋅(79)=−1(4−h−2​)⋅(97​)=−1 −149(4−ℎ)=−19(4−h)−14​=−1

    4. Solve for ℎh: −149(4−ℎ)=−19(4−h)−14​=−1 149(4−ℎ)=19(4−h)14​=1 14=9(4−ℎ)14=9(4−h) 14=36−9ℎ14=36−9h 9ℎ=36−149h=36−14 9ℎ=229h=22 ℎ=229h=922​

    So, the value of ℎh is 229922​.

  12. 12.Exercise Questions

    Question 1

    The perpendicular from the origin to the line 𝑦=𝑚𝑥+𝑐y=mx+c meets it at the point (−1,2)(−1,2). Find the values of 𝑚m and 𝑐c.

    Solution

    1. The equation of the line: Given the line equation 𝑦=𝑚𝑥+𝑐y=mx+c.

    2. The point of intersection: The perpendicular from the origin (0, 0) to this line meets the line at the point (−1,2)(−1,2).

    3. Slope of the perpendicular line: The slope of the given line is 𝑚m, so the slope of the perpendicular line is −1𝑚−m1​.

    4. Equation of the perpendicular line: The line passing through the origin with slope −1𝑚−m1​ is: 𝑦=−1𝑚𝑥y=−m1​x

    5. Point of intersection: The point (−1,2)(−1,2) lies on both lines. Therefore: 2=𝑚(−1)+𝑐2=m(−1)+c 2=−𝑚+𝑐2=−m+c 𝑐=2+𝑚c=2+m

      Also, substituting (−1,2)(−1,2) in the perpendicular line equation: 2=−1𝑚(−1)2=−m1​(−1) 2=1𝑚2=m1​ 𝑚=12m=21​

    6. Find 𝑐c: Substituting 𝑚=12m=21​ into 𝑐=2+𝑚c=2+m: 𝑐=2+12c=2+21​ 𝑐=2.5c=2.5

    So, the values are 𝑚=12m=21​ and 𝑐=2.5c=2.5.

    Question 2

    If 𝑝p and 𝑞q are the lengths of perpendiculars from the origin to the lines 𝑥cos⁡𝜃−𝑦sin⁡𝜃=𝑘cos⁡2𝜃xcosθ−ysinθ=kcos2θ and 𝑥sec⁡𝜃+𝑦csc⁡𝜃=𝑘xsecθ+ycscθ=k, respectively, prove that 𝑝2+4𝑞2=𝑘2p2+4q2=k2.

    Solution

    1. Length of perpendicular from the origin to the line 𝐴𝑥+𝐵𝑦+𝐶=0Ax+By+C=0: Length=∣𝐶∣𝐴2+𝐵2Length=A2+B2​∣C∣​

    2. For the first line 𝑥cos⁡𝜃−𝑦sin⁡𝜃=𝑘cos⁡2𝜃xcosθ−ysinθ=kcos2θ:

      • Here, 𝐴=cos⁡𝜃A=cosθ, 𝐵=−sin⁡𝜃B=−sinθ, 𝐶=−𝑘cos⁡2𝜃C=−kcos2θ.
      • Length 𝑝p: 𝑝=∣−𝑘cos⁡2𝜃∣(cos⁡𝜃)2+(−sin⁡𝜃)2p=(cosθ)2+(−sinθ)2​∣−kcos2θ∣​ 𝑝=𝑘cos⁡2𝜃cos⁡2𝜃+sin⁡2𝜃p=cos2θ+sin2θ​kcos2θ​ 𝑝=𝑘cos⁡2𝜃p=kcos2θ

    3. For the second line 𝑥sec⁡𝜃+𝑦csc⁡𝜃=𝑘xsecθ+ycscθ=k:

      • Here, 𝐴=sec⁡𝜃A=secθ, 𝐵=csc⁡𝜃B=cscθ, 𝐶=−𝑘C=−k.
      • Length 𝑞q: 𝑞=∣−𝑘∣(sec⁡𝜃)2+(csc⁡𝜃)2q=(secθ)2+(cscθ)2​∣−k∣​ 𝑞=𝑘sec⁡2𝜃+csc⁡2𝜃q=sec2θ+csc2θ​k​ sec⁡2𝜃=1+tan⁡2𝜃sec2θ=1+tan2θ csc⁡2𝜃=1+cot⁡2𝜃csc2θ=1+cot2θ 𝑞=𝑘1+tan⁡2𝜃+1+cot⁡2𝜃q=1+tan2θ+1+cot2θ​k​ 𝑞=𝑘2+tan⁡2𝜃+cot⁡2𝜃q=2+tan2θ+cot2θ​k​

    4. Prove that 𝑝2+4𝑞2=𝑘2p2+4q2=k2: 𝑝=𝑘cos⁡2𝜃p=kcos2θ 𝑞=𝑘2+tan⁡2𝜃+cot⁡2𝜃q=2+tan2θ+cot2θ​k​

      • We know that cos⁡2𝜃=cos⁡2𝜃−sin⁡2𝜃cos2θ=cos2θ−sin2θ.

    Question 3

    In the triangle ABC with vertices A(2, 3), B(4, -1), and C(1, 2), find the equation and length of altitude from the vertex A.

    Solution

    1. Equation of line BC: Slope of BC=2+11−4=3−3=−1Slope of BC=1−42+1​=−33​=−1 Equation of BC=𝑦−(−1)=−1(𝑥−4)Equation of BC=y−(−1)=−1(x−4) 𝑦+1=−𝑥+4y+1=−x+4 𝑥+𝑦=3x+y=3

    2. Equation of altitude from A:

      • Slope of altitude from A is the negative reciprocal of slope of BC: 1 𝑦−3=1(𝑥−2)y−3=1(x−2) 𝑦−3=𝑥−2y−3=x−2 𝑦=𝑥+1y=x+1

    3. Length of altitude: Length=∣2⋅1+3⋅1−3∣12+12Length=12+12​∣2⋅1+3⋅1−3∣​ =∣2+3−3∣2=2​∣2+3−3∣​ =22=2=2​2​=2​

    Question 4

    If 𝑝p is the length of perpendicular from the origin to the line whose intercepts on the axes are 𝑎a and 𝑏b, then show that 1𝑝2=1𝑎2+1𝑏2p21​=a21​+b21​.

    Solution

    1. Line equation in intercept form: 𝑥𝑎+𝑦𝑏=1ax​+by​=1

      • Rewrite in standard form: 𝑏𝑥+𝑎𝑦=𝑎𝑏bx+ay=ab.

    2. Length of perpendicular from origin: 𝑝=∣𝑎𝑏∣𝑎2+𝑏2p=a2+b2​∣ab∣​

    3. Square both sides: 𝑝2=𝑎2𝑏2𝑎2+𝑏2p2=a2+b2a2b2​ 1𝑝2=𝑎2+𝑏2𝑎2𝑏2p21​=a2b2a2+b2​ 1𝑝2=𝑎2𝑎2𝑏2+𝑏2𝑎2𝑏2p21​=a2b2a2​+a2b2b2​ 1𝑝2=1𝑏2+1𝑎2p21​=b21​+a21​

    Thus, 1𝑝2=1𝑎2+1𝑏2p21​=a21​+b21​.

More Class 11 Maths chapters