Chapter 8: Coordinate Geometry
Straight-line equations, slope, angle between two lines, parallel and perpendicular conditions, and applications — reconstructed from the supplied 89-page handwritten notes.
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Basic Coordinate Geometry
Equation of Axes
Lines Parallel to Axes
Forms of the Equation of a Straight Line
m = slope, c = y-intercept
a = x-intercept, b = y-intercept
p = perpendicular distance from origin
Theory 1: Angle Between Two Straight Lines
If two lines have slopes m₁ and m₂, and θ is the angle between them, then:
Derivation Used in the Source
The notes consider the inclination angles θ₁ and θ₂ of the two lines and use the tangent subtraction formula:
Since tan θ₁ = m₁ and tan θ₂ = m₂, the required relation follows.
Conditions for Parallel and Perpendicular Lines
Perpendicular Condition
Parallel Condition
For Lines a₁x+b₁y+c₁=0 and a₂x+b₂y+c₂=0
Line Parallel to ax+by+c=0
Line Perpendicular to ax+by+c=0
Exercise 8.1
Very Short Type Questions
1. Angle between y=m₁x+c₁ and y=m₂x+c₂
Slopes are m₁ and m₂.
Therefore:
θ = tan⁻¹[±(m₁−m₂)/(1+m₁m₂)]
2. Conditions for parallelism and perpendicularity
Parallel: m₁=m₂
Perpendicular: m₁m₂=−1
3. Find the slope of 2x+3y=6
m = −x-coefficient / y-coefficient
= −2/3
∴ slope = −2/3
4. Midpoint and slope joining (2,3) and (4,5)
Midpoint = ((2+4)/2, (3+5)/2)
∴ (3,4)
Slope = (5−3)/(4−2)=1
∴ slope = 1
5. Slopes parallel/perpendicular to a given line
Pages 13–14 take the given straight line, calculate its slope, then use:
Short Questions: Acute and Obtuse Angles Between Lines
2y+5=0 and √3y−3x=0
m₁=0, m₂=√3
tan θ = ±(0−√3)/(1+0)
|tan θ|=√3
∴ acute angle = 60°
x−2y+7=0 and y=5−x
m₁=1/2, m₂=−1
The source substitutes in the angle formula and leaves the result in inverse-tangent form.
y=(2−√3)x+6 and y=(2+√3)x+2
m₁=2−√3, m₂=2+√3
tan θ = √3
∴ θ=60°
y=√3x+4 and x−√3y=5
m₁=√3
m₂=1/√3
tan θ=1/√3
∴ θ=30°
y+6=0 and √3x+y=5
m₁=0, m₂=−√3
The source selects the obtuse angle:
∴ θ=120°
x+(3/2)y=−2 and 3x+2y=4
m₁=−2/3, m₂=−3/2
The notes evaluate the angle using the standard formula.
y−(2+√3)x=7 and y−(2−√3)x=6
The source obtains an obtuse angle:
∴ θ=120°
2x−y+4=0 and 3x+y+3=0
m₁=2, m₂=−3
|tan θ|=1
∴ θ=135° for the obtuse angle shown in the source.
Angles Between Each Pair of Straight Lines
y−√3x+8=0 and y+10=0
m₁=√3, m₂=0
tan θ=±√3
Hence the source records:
θ=60° or 120°
3x−y+4=0 and 2x+y+3=0
m₁=3, m₂=−2
tan θ=±1
∴ θ=45° or 135°
x−√3y=a and √3x−y=b
m₁=1/√3, m₂=√3
tan θ=±1/√3
∴ θ=30° or 150°
x sec θ + y cosec θ = a and x cos θ − y sin θ = b
The source derives slopes −tan θ and cot θ.
Their product is −1.
∴ the lines are perpendicular and the angle is 90°
Show that the Following Pairs of Lines are Parallel
The source repeatedly calculates the slopes of the two lines and verifies that:
(a) For the first pair, both slopes are −1.
(b) For another pair, both slopes are 1/4.
(c) The next pair again has equal slopes and is therefore parallel.
(d) Another pair is shown parallel by identical slope values.
Prove that the Following Pairs of Lines are Perpendicular
For each pair, the notes calculate m₁ and m₂ and verify:
(a) m₁ = 2/3, m₂ = −3/2
(b) m₁ = −3/2, m₂ = 2/3
(c) Another pair gives reciprocal slopes with opposite signs.
(d) The final pair is also verified using m₁m₂=−1.
Finding Unknown Parameters for Parallel or Perpendicular Lines
If ax+3y−8=0 and 2y−3x−11=0 are parallel
m₁=−a/3, m₂=3/2
Since m₁=m₂:
−a/3 = 3/2
∴ a = −9/2
If ax−3y=0 and 3y+x=3 are parallel
The source equates slopes and obtains:
∴ a = −1/3
Parameter k for perpendicular lines
Pages 42–46 contain several questions where one coefficient is k. The method is:
One problem gives k = −4.
Another gives k = 4.
Another gives k = −8/3.
Line kx−3y+6=0 perpendicular to line joining two points
The slope of the joining line is first found from the two points.
Using m₁m₂=−1, the source obtains:
∴ k = 1/4
Line ax+3y+5=0 perpendicular to line joining two points
The slope of the joining line is −3.
Using perpendicularity:
−a/3 × (−3) = −1
∴ a = −1
Find slope of a line perpendicular to ax+3y=12
Slope of given line = −a/3.
Let perpendicular slope be m₂.
(−a/3)m₂=−1
∴ m₂ = 3/a
Long Type Questions
Equations of Lines Parallel to Given Lines
Through a point and parallel to a line joining two points
Pages 48–52 use the two-point slope formula to find the slope of a reference line, then use the point-slope form for a parallel line through the required point.
One worked result is:
x + 4y − 9 = 0
Another worked result
The source obtains:
2x + 5y − 29 = 0
Line parallel to 5x+4y−9=0 with x-intercept −5
Given slope = −5/4.
Using point (−5,0):
y = −5/4(x+5)
∴ 5x + 4y + 25 = 0
Line parallel to 3x+4y=12 through midpoint of (2,6) and (4,2)
Midpoint = (3,4)
Given slope = −3/4
Using point-slope form:
∴ 3x + 4y − 25 = 0
Equations of Lines Perpendicular to Given Lines
Through a point and perpendicular to 6x+5y+14=0
Given slope = −6/5.
Perpendicular slope = 5/6.
Using the specified point, the source obtains:
5x − 6y + 16 = 0
Perpendicular to 5x+7y=12 through (2,−3)
Given slope = −5/7.
Perpendicular slope = 7/5.
Using point-slope form:
∴ 7x − 5y − 29 = 0
Perpendicular to 12x−5y+82=0 through (−3,−9)
Given slope = 12/5.
Perpendicular slope = −5/12.
Source result:
5x + 12y + 63 = 0
Perpendicular to 3x−y=2 and making y-intercept 4
Given slope = 3.
Perpendicular slope = −1/3.
Using (0,4):
∴ x + 3y − 12 = 0
Equation of the Perpendicular Bisector of a Line Segment
The method used repeatedly in the source is:
1. Find the midpoint of the two given points.
2. Find the slope of the line segment.
3. Use m₁m₂=−1 to find the perpendicular slope.
4. Use the point-slope form through the midpoint.
For points (5,4) and (7,12)
Midpoint = (6,8)
Slope of joining line = 4
Perpendicular slope = −1/4
Source result:
∴ x + 4y − 38 = 0
For points (3,−7) and (−5,3)
Midpoint = (−1,−2)
Slope of joining line = −5/4
Perpendicular slope = 4/5
Source result:
∴ 4x − 5y − 6 = 0
For points (4,−2) and (6,2)
Midpoint = (5,0)
Slope = 2
Perpendicular slope = −1/2
∴ x + 2y − 5 = 0
Another segment problem
Pages 71–73 repeat the midpoint-and-perpendicular-slope method for another pair of points and obtain a straight-line equation in standard form.
Midpoint and Perpendicular-Line Applications
Coordinates A(2,−1) and B(3,3)
Midpoint = (5/2,1)
Slope AB = 4
Perpendicular slope = −1/4
The source forms the required perpendicular line through the midpoint.
Coordinates A(3,−1) and B(7,1)
Midpoint = (5,0)
Slope AB = 1/2
Perpendicular slope = −2
∴ 2x + y − 10 = 0
Square and Rhombus Diagonal Problems
Square: endpoints of one diagonal are (2,3) and (−6,5)
Midpoint of the given diagonal = (−2,4)
Slope of the given diagonal = −1/4
Since diagonals of a square are perpendicular, slope of other diagonal = 4.
The source writes the equation of the second diagonal through the midpoint.
Rhombus with opposite vertices A(2,2) and C(8,10)
Midpoint of AC = (5,6)
Slope of AC = 1
Diagonals of a rhombus are perpendicular.
Hence slope of BD = −1.
The source obtains the equation of BD through (5,6).
Triangle Altitude Problems
Triangle ABC with vertices approximately A(2,3), B(8,10), C(−2,−1)
The source first finds the midpoint of a side and then forms a required line from a vertex to that midpoint.
Altitude from A(2,3) to BC
Slope of BC = −1/6
Altitude is perpendicular to BC, so its slope = 6.
Using point (2,3):
y−3=6(x−2)
∴ 6x − y − 9 = 0
Lines Through a Point Making 45° with a Given Line
The source uses the angle formula with tan 45°=1:
Through (2,3) and making 45° with 2x+3y=2
Slope of the given line = −2/3.
Solving the angle condition gives two possible slopes:
m = 1/5 and m = −5
Using y−3=m(x−2):
For m=1/5, the source obtains x − 5y + 13 = 0.
For m=−5, the source obtains 5x + y − 13 = 0.
Through (1,−4) and making 45° with 2x+3y+5=0
The source again solves the angle formula and obtains two slopes:
m = 1/5 and m = −5
Corresponding lines through (1,−4) are formed using point-slope form.
Discussion
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