Grade 10 Notes of C0-ORDINATE GEOMETRY || Optional Mathematics

1.Introduction to Coordinate Geometry

Coordinate Geometry connects algebra and geometry, allowing us to represent points, lines, and curves using algebraic equations.

1.1.Cartesian Plane
1.X-axis → horizontal line
2.Y-axis → vertical line
3.Origin → intersection of X and Y axes (0,0)

1.2.Quadrants

1.3.Distance Formula
Distance between A(x₁,y₁) and B(x₂,y₂):
AB = √[(x₂−x₁)² + (y₂−y₁)²]

1.4.Midpoint / Section Formula
Midpoint of AB = ((x₁+x₂)/2 , (y₁+y₂)/2)

2.Straight Lines

A straight line is the shortest path between two points.

2.1.Slope of a Line
Slope (m) = (y₂ − y₁) / (x₂ − x₁)

1.m > 0 → line rises
2.m < 0 → line falls
3.m = 0 → horizontal line
4.m undefined → vertical line

2.2.Equation of Line

1.Point-Slope Form: y−y₁ = m(x−x₁)
2.Two-Point Form: y−y₁ = (y₂−y₁)/(x₂−x₁)
3.Slope-Intercept Form: y = mx + c
4.General Form: ax + by + c = 0 (slope = −a/b)

2.3.Conditions for Lines

1.Parallel → m₁ = m₂
2.Perpendicular → m₁ × m₂ = −1
3.Horizontal → y = constant
4.Vertical → x = constant

3.Angle Between Two Straight Lines

3.1.Slope and Inclination
1.m = tanθ, where θ is angle with positive X-axis

3.2.Angle Between Two Lines (Slopes m₁ and m₂)

tanθ = |(m₁ − m₂) / (1 + m₁m₂)|

1.Absolute value gives acute angle
2.± can be used to indicate both acute/obtuse angles

Special Cases
1.Parallel → m₁ = m₂ → θ = 0° or 180°
2.Perpendicular → m₁ × m₂ = −1

3.3.Angle Between Lines in General Form

1.Lines: a₁x + b₁y + c₁ = 0, a₂x + b₂y + c₂ = 0
2.Slopes: m₁ = −a₁/b₁, m₂ = −a₂/b₂
3.Angle: tanθ = |(a₁b₂ − a₂b₁)/(a₁a₂ + b₁b₂)|

Special Cases
1.Parallel → a₁/a₂ = b₁/b₂ (c₁/c₂ ≠ same)
2.Perpendicular → a₁a₂ + b₁b₂ = 0

3.4.Applications

1.Parallel line through point → same slope
2.Perpendicular line through point → slope = −1/m
3.Line at a given angle → use tanθ formula
4.Perpendicular bisector → midpoint + negative reciprocal slope
5.Diagonals of square → intersect at midpoint and perpendicular

4.Pair of Straight Lines

Equation: ax² + 2hxy + by² = 0

1.Represents two straight lines passing through origin
2.Angle between lines: tanθ = |2√(h² − ab)/(a + b)|

Conditions
1.Perpendicular → a + b = 0
2.Parallel/Coincident → h² = ab

Applications
1.Find combined equation of lines
2.Find perpendicular lines
3.Solve problems involving angles

5.Conic Sections (Introduction)

1.Curves obtained by cutting a double cone with a plane

5.1.Types of Conics

• Right Circular Cone: Vertex O, axis (vertical), generator (slant), base.
• Sections:
  • Circle: Plane || base.
  • Parabola: Plane || generator.
  • Ellipse: Plane angle > vertical angle but <90°.
  • Hyperbola: Plane || axis (double cone).
  • 

6.Circle

6.1.Definition
A circle is the locus of a point moving at a fixed distance r from a fixed centre.

6.2.Equations of Circle

1.Centre (0,0): x² + y² = r²
2.Centre (h,k): (x−h)² + (y−k)² = r²
3.General Form: x² + y² + 2gx + 2fy + c = 0
Centre = (−g,−f), radius = √(g² + f² − c)

6.3.Special Cases

1.Circle touches x-axis → centre (h,r)
2.Circle touches y-axis → centre (r,k)
3.Circle passes through origin → (x−a)² + (y−b)² = r²
4.Angle in semicircle = 90°
5.Four points on same circle → concyclic

7.Important Formulas 

8.Important Questions

Question 1
Find the equation of the line passing through (2, 3) and parallel to
4x - 5y + 10 = 0.

Solution:
Given line: 4x - 5y + 10 = 0
Rewrite in slope form:

5y = 4x + 10
y = (4/5)x + 2

So slope m = 4/5

For a parallel line, slope remains same.
Using point-slope form:

y - 3 = (4/5)(x - 2)

Multiply both sides by 5:

5y - 15 = 4x - 8

Rearranging:

4x - 5y + 7 = 0

Question 2
Find the acute angle between the lines
3x - 4y + 5 = 0 and 4x + 3y - 1 = 0.

Solution:
First line:

3x - 4y + 5 = 0
4y = 3x + 5
y = (3/4)x + 5/4

So slope m1 = 3/4

Second line:

4x + 3y - 1 = 0
3y = -4x + 1
y = (-4/3)x + 1/3

So slope m2 = -4/3

Formula for angle between two lines:

tan(theta) = |(m1 - m2) / (1 + m1m2)|

Substitute values:

tan(theta) = |(3/4 + 4/3) / (1 - 1)|

Denominator becomes zero.
Therefore theta = 90 degrees.

So the acute angle between the lines is 90 degrees.

Question 3
Find the combined equation of the pair of lines
x - y = 0 and 2x + y = 0.

Solution:
Combined equation is product of both equations:

(x - y)(2x + y) = 0

Multiply:

2x squared + xy - 2xy - y squared = 0

Simplify:

2x squared - xy - y squared = 0

Question 4
Find the equation of the circle with centre (3, -2) and radius 5.

Solution:
Standard form of circle:

(x - h) squared + (y - k) squared = r squared

Here h = 3, k = -2, r = 5

Substitute:

(x - 3) squared + (y + 2) squared = 25

Expand:

x squared - 6x + 9 + y squared + 4y + 4 = 25

Rearrange:

x squared + y squared - 6x + 4y - 12 = 0

Question 5
Find the centre and radius of the circle
x squared + y squared + 4x - 6y - 12 = 0.

Solution:
Group x and y terms:

(x squared + 4x) + (y squared - 6y) = 12

Complete square:

(x + 2) squared - 4 + (y - 3) squared - 9 = 12

Add constants:

(x + 2) squared + (y - 3) squared = 25

So,
Centre = (-2, 3)
Radius = 5

MORE PRACTICE QUESTIONS (MODERATE TO COMPLEX)

Question 6
Find equation of the line passing through (1, -2) and perpendicular to
2x + 3y - 7 = 0.

Solution:
Given line:

2x + 3y - 7 = 0
3y = -2x + 7
y = (-2/3)x + 7/3

Slope of given line = -2/3

Slope of perpendicular line = 3/2

Using point-slope form:

y + 2 = (3/2)(x - 1)

Multiply by 2:

2y + 4 = 3x - 3

Rearrange:

3x - 2y - 7 = 0

Question 7
Find the angle between lines
x + y = 0 and x - y = 0.

Solution:
First line slope = -1
Second line slope = 1

tan(theta) = |(-1 - 1) / (1 - 1)|

Denominator zero.
So angle = 90 degrees.

Question 8
Find combined equation of pair of lines passing through origin and having slopes 2 and -3.

Solution:
General equation:

(y - 2x)(y + 3x) = 0

Multiply:

y squared + 3xy - 2xy - 6x squared = 0

Simplify:

y squared + xy - 6x squared = 0

Question 9
Find the equation of circle passing through (0,0) with centre on x-axis and radius 5.

Solution:
Let centre be (a, 0).

Using circle equation:

(x - a) squared + y squared = 25

Since circle passes through origin:

a squared = 25

So a = 5 or a = -5

Hence equations:

(x - 5) squared + y squared = 25
or
(x + 5) squared + y squared = 25

Question 10
Find equation of circle passing through points (1,0), (0,1), and (0,-1).

Solution:
General circle equation:

x squared + y squared + 2gx + 2fy + c = 0

Substitute points one by one.

For (1,0):

1 + 0 + 2g + c = 0

For (0,1):

1 + 2f + c = 0

For (0,-1):

1 - 2f + c = 0

Subtract last two equations:

4f = 0
So f = 0

Then c = -1

Then 2g = 0
So g = 0

Final equation:

x squared + y squared - 1 = 0

Visit this link for further practice!!

https://besidedegree.com/exam/s/academic