Coordinate Geometry

Master the art of plotting points, lines, and curves in the coordinate plane. Learn about distance, slope, equations, and their applications.

Table of Contents

Simplified Quantitative Formulas: Coordinate Geometry

  • Distance Formula: √[(x₂–x₁)² + (y₂–y₁)²]
  • Midpoint Formula: ((x₁+x₂)/2, (y₁+y₂)/2)
  • Slope (m): (y₂–y₁)/(x₂–x₁)
  • Equation of Line: y = mx + c or (y–y₁) = m(x–x₁)
  • Section Formula: (mx₂+nx₁)/(m+n), (my₂+ny₁)/(m+n)
  • Area of Triangle: (1/2)|x₁(y₂–y₃) + x₂(y₃–y₁) + x₃(y₁–y₂)|
  • Circle: (x–h)² + (y–k)² = r²
  • Variable Definitions: (x₁, y₁), (x₂, y₂), (x₃, y₃) = points; m, n = ratios; h, k = center; r = radius

What do these mean? (Super Simple Explanations & Examples)

  • Distance: How far between two points. Example: (1,2) to (4,6): √[(4–1)² + (6–2)²] = √(9+16) = 5.
  • Midpoint: Middle point. Example: (2,3) & (4,7): ((2+4)/2, (3+7)/2) = (3,5).
  • Slope: Steepness of a line. Example: (1,2) & (3,6): (6–2)/(3–1) = 2.
  • Equation of Line: y = mx + c. Example: Slope = 2, y-intercept = 1: y = 2x + 1.
  • Area of Triangle: Use formula with 3 points. Example: (0,0), (4,0), (0,3): Area = (1/2)|0(0–3)+4(3–0)+0(0–0)| = 6.
  • Circle: (x–2)² + (y–3)² = 25 is a circle with center (2,3), radius 5.

1. Introduction to Coordinate Geometry

Coordinate geometry is a branch of mathematics that combines algebra and geometry. It allows us to represent geometric shapes using algebraic equations and vice versa.

Key Components:

  • Coordinate System: A system for specifying points using ordered pairs (x, y)
  • X-axis: The horizontal number line
  • Y-axis: The vertical number line
  • Origin: The point (0,0) where axes intersect
  • Quadrants: Four regions created by the axes
Coordinate System

Key Concepts & Properties

Quadrants in the Coordinate Plane

Quadrant Sign of X Sign of Y
IPositivePositive
IINegativePositive
IIINegativeNegative
IVPositiveNegative

The coordinate plane is divided into four quadrants by the x- and y-axes. The sign of x and y determines the quadrant in which a point lies.

Collinear Points

Three or more points are collinear if they all lie on the same straight line. To check if points A(x₁, y₁), B(x₂, y₂), and C(x₃, y₃) are collinear, calculate the area of the triangle they form. If the area is zero, the points are collinear.

Formula: Area = ½ |x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂)| = 0 ⇒ Collinear

Concurrent Lines

Three or more lines are concurrent if they all pass through a single point (called the point of concurrency). For example, the medians of a triangle are concurrent at the centroid.

Midpoint Formula

The midpoint of a segment joining A(x₁, y₁) and B(x₂, y₂) is:

Midpoint = ( (x₁ + x₂)/2 , (y₁ + y₂)/2 )

Section Formula (External Division)

If a point P divides the line AB externally in the ratio m:n, then:

P = ( (mx₂ - nx₁)/(m-n), (my₂ - ny₁)/(m-n) )

Incentre of a Triangle

The incentre is the point where the angle bisectors of a triangle meet. For triangle ABC with vertices A(x₁, y₁), B(x₂, y₂), C(x₃, y₃) and side lengths a (BC), b (AC), c (AB):

Incentre = ( (a·x₁ + b·x₂ + c·x₃)/(a+b+c), (a·y₁ + b·y₂ + c·y₃)/(a+b+c) )

Perpendicular Distance from a Point to a Line

For point (x₁, y₁) and line Ax + By + C = 0:

Distance = |A·x₁ + B·y₁ + C| / √(A² + B²)

Distance Between Two Parallel Lines

For lines Ax + By + C₁ = 0 and Ax + By + C₂ = 0:

Distance = |C₁ - C₂| / √(A² + B²)

Parallel Lines and Transversals

  • When a transversal crosses parallel lines, corresponding angles are equal.
  • Alternate interior angles are equal.
  • Interior angles on the same side of the transversal are supplementary (sum to 180°).
  • The sum of all angles at a point where lines meet is 360°.

Angle Properties

  • Complementary angles: Two angles whose sum is 90°.
  • Supplementary angles: Two angles whose sum is 180°.
  • When two lines intersect, opposite (vertical) angles are equal.
  • Adjacent angles formed by intersecting lines are supplementary.

Ratio Theorem for Transversals and Parallel Lines

If two transversals are cut by three or more parallel lines, the segments intercepted on one transversal are proportional to the corresponding segments on the other transversal.

If PR/RT = QS/SU, then the lines are parallel and the segments are in the same ratio.

2. Basic Concepts

(a) Distance Formula

For points P(x₁, y₁) and Q(x₂, y₂):

d = √[(x₂ - x₁)² + (y₂ - y₁)²]

Distance Between Points

(b) Section Formula

For point P dividing line segment AB in ratio m:n:

P = ((mx₂ + nx₁)/(m+n), (my₂ + ny₁)/(m+n))

(c) Area of Triangle

For points A(x₁, y₁), B(x₂, y₂), and C(x₃, y₃):

Area = ½|(x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂))|

3. Lines and Their Equations

(a) Slope of a Line

For points P(x₁, y₁) and Q(x₂, y₂):

m = (y₂ - y₁)/(x₂ - x₁)

(b) Forms of Line Equations

  • Point-Slope Form: y - y₁ = m(x - x₁)
  • Slope-Intercept Form: y = mx + c
  • Two-Point Form: (y - y₁)/(y₂ - y₁) = (x - x₁)/(x₂ - x₁)
  • Intercept Form: x/a + y/b = 1

(c) Angle Between Lines

For lines with slopes m₁ and m₂:

tan θ = |(m₁ - m₂)/(1 + m₁m₂)|

4. Circles

(a) Standard Equation

For circle with center (h,k) and radius r:

(x - h)² + (y - k)² = r²

(b) General Equation

x² + y² + 2gx + 2fy + c = 0

where center = (-g,-f) and radius = √(g² + f² - c)

5. Conic Sections

(a) Parabola

Standard form: y² = 4ax

Vertex at origin, focus at (a,0)

(b) Ellipse

Standard form: x²/a² + y²/b² = 1

Center at origin, foci at (±c,0) where c² = a² - b²

(c) Hyperbola

Standard form: x²/a² - y²/b² = 1

Center at origin, foci at (±c,0) where c² = a² + b²

6. Transformations

(a) Translation

Point (x,y) → (x+h, y+k)

(b) Rotation

Point (x,y) → (xcosθ - ysinθ, xsinθ + ycosθ)

(c) Reflection

Over x-axis: (x,y) → (x,-y)

Over y-axis: (x,y) → (-x,y)

7. Applications and Practice Questions

Basic Concepts Questions

Question 1
Find the distance between points A(3,4) and B(7,1).
Solution:
1. Using distance formula:
2. d = √[(7-3)² + (1-4)²]
3. = √[16 + 9]
4. = √25 = 5 units
Question 2
Find the point that divides the line segment joining A(2,3) and B(6,7) in the ratio 2:1.
Solution:
1. Using section formula:
2. P = ((2×6 + 1×2)/(2+1), (2×7 + 1×3)/(2+1))
3. = ((12 + 2)/3, (14 + 3)/3)
4. = (14/3, 17/3)
Question 3
Find the area of triangle with vertices A(0,0), B(4,0), and C(2,3).
Solution:
1. Using area formula:
2. Area = ½|(0(0-3) + 4(3-0) + 2(0-0))|
3. = ½|(0 + 12 + 0)|
4. = 6 square units
Question 4
Find the coordinates of the centroid of triangle with vertices A(1,2), B(4,6), and C(7,4).
Solution:
1. Centroid coordinates = ((1+4+7)/3, (2+6+4)/3)
2. = (12/3, 12/3)
3. = (4, 4)

Lines and Equations Questions

Question 1
Find the equation of a line passing through points (2,3) and (4,7).
Solution:
1. First find slope: m = (7-3)/(4-2) = 2
2. Using point-slope form:
3. y - 3 = 2(x - 2)
4. y = 2x - 1
Question 2
Find the angle between lines with slopes 2 and -1/2.
Solution:
1. Using angle formula:
2. tan θ = |(2 - (-1/2))/(1 + 2(-1/2))|
3. = |(5/2)/0| = undefined
4. Therefore, θ = 90°
Question 3
Find the equation of a line parallel to 2x + 3y = 6 and passing through (1,2).
Solution:
1. Slope of given line = -2/3
2. Parallel line has same slope
3. Using point-slope form:
4. y - 2 = (-2/3)(x - 1)
5. 3y - 6 = -2x + 2
6. 2x + 3y = 8
Question 4
Find the perpendicular distance from point (3,4) to line 3x + 4y = 10.
Solution:
1. Using distance formula:
2. d = |3(3) + 4(4) - 10|/√(3² + 4²)
3. = |9 + 16 - 10|/5
4. = 15/5 = 3 units

Circle Questions

Question 1
Find the center and radius of the circle x² + y² - 6x + 8y = 0.
Solution:
1. Complete the square:
2. (x² - 6x + 9) + (y² + 8y + 16) = 25
3. (x-3)² + (y+4)² = 5²
4. Center = (3,-4), Radius = 5
Question 2
Find the equation of circle passing through points (1,2), (3,4), and (5,2).
Solution:
1. Let equation be x² + y² + 2gx + 2fy + c = 0
2. Substitute points to get three equations
3. Solve to get g = -3, f = -3, c = 8
4. Final equation: x² + y² - 6x - 6y + 8 = 0
Question 3
Find the length of tangent from point (4,3) to circle x² + y² = 25.
Solution:
1. Using tangent length formula:
2. Length = √(4² + 3² - 25)
3. = √(16 + 9 - 25)
4. = 0 (point lies on circle)
Question 4
Find the equation of circle touching x-axis at (3,0) and passing through (1,2).
Solution:
1. Center lies on x = 3
2. Let center be (3,k)
3. Radius = k (touches x-axis)
4. (1-3)² + (2-k)² = k²
5. 4 + 4 - 4k + k² = k²
6. k = 2
7. Equation: (x-3)² + (y-2)² = 4

Conic Sections Questions

Question 1
Find the focus and directrix of parabola y² = 8x.
Solution:
1. Compare with y² = 4ax
2. 4a = 8, so a = 2
3. Focus = (2,0)
4. Directrix: x = -2
Question 2
Find the foci and vertices of ellipse 9x² + 16y² = 144.
Solution:
1. Divide by 144: x²/16 + y²/9 = 1
2. a = 4, b = 3
3. c = √(16-9) = √7
4. Foci = (±√7,0)
5. Vertices = (±4,0)
Question 3
Find the asymptotes of hyperbola x²/9 - y²/16 = 1.
Solution:
1. Compare with x²/a² - y²/b² = 1
2. a = 3, b = 4
3. Asymptotes: y = ±(b/a)x
4. y = ±(4/3)x
Question 4
Find the equation of parabola with vertex at (2,3) and focus at (4,3).
Solution:
1. Vertex (h,k) = (2,3)
2. Focus (h+a,k) = (4,3)
3. So a = 2
4. Equation: (y-3)² = 8(x-2)

Transformations Questions

Question 1
Find the image of point (3,4) after translation by vector (2,-1).
Solution:
1. New x = 3 + 2 = 5
2. New y = 4 + (-1) = 3
3. Image point = (5,3)
Question 2
Find the image of point (2,3) after rotation of 90° about origin.
Solution:
1. For 90° rotation:
2. x' = -y = -3
3. y' = x = 2
4. Image point = (-3,2)
Question 3
Find the image of point (4,5) after reflection over line y = x.
Solution:
1. For reflection over y = x:
2. x' = y = 5
3. y' = x = 4
4. Image point = (5,4)
Question 4
Find the image of point (3,2) after dilation with scale factor 2 about origin.
Solution:
1. For dilation with scale factor k:
2. x' = kx = 2×3 = 6
3. y' = ky = 2×2 = 4
4. Image point = (6,4)

Practice Questions: Coordinate Geometry

Test your understanding of all key coordinate geometry concepts with these original, exam-style questions. Click each question to reveal the answer and explanation!

Practice More: Coordinate Geometry →
1. In which quadrant does the point (-3, 5) lie?
Answer & Explanation:
The x-coordinate is negative and the y-coordinate is positive, so the point lies in Quadrant II.
2. Are the points A(1, 2), B(3, 6), and C(5, 10) collinear?
Answer & Explanation:
Calculate the area: ½|1(6-10) + 3(10-2) + 5(2-6)| = ½|-4 + 24 - 20| = ½|0| = 0. Since the area is zero, the points are collinear.
3. Three lines x + y = 2, x - y = 0, and y = 1 are drawn. Do they all meet at a single point?
Answer & Explanation:
Solve x - y = 0 and y = 1: x = 1, y = 1. Substitute into x + y = 2: 1 + 1 = 2. All three lines meet at (1, 1), so they are concurrent.
4. What is the midpoint of the segment joining (2, 7) and (6, 3)?
Answer & Explanation:
Midpoint = ((2+6)/2, (7+3)/2) = (4, 5).
5. Find the point that divides the segment from (1, 2) to (7, 8) externally in the ratio 2:1.
Answer & Explanation:
P = ((2×7 - 1×1)/(2-1), (2×8 - 1×2)/(2-1)) = (13/1, 14/1) = (13, 14).
6. For triangle A(0,0), B(6,0), C(0,8), find the coordinates of the incentre.
Answer & Explanation:
First, find side lengths: a = BC = 10, b = AC = 8, c = AB = 6. Incentre = ((a×0 + b×6 + c×0)/(a+b+c), (a×0 + b×0 + c×8)/(a+b+c)) = (48/24, 48/24) = (2, 2).
7. What is the perpendicular distance from (3, 4) to the line 5x + 12y - 13 = 0?
Answer & Explanation:
Distance = |5×3 + 12×4 - 13| / √(25+144) = |15+48-13|/13 = 50/13 ≈ 3.85 units.
8. What is the distance between the lines 3x - 4y + 7 = 0 and 3x - 4y - 5 = 0?
Answer & Explanation:
Distance = |7 - (-5)| / √(9+16) = 12/5 = 2.4 units.
9. In the figure, two parallel lines are cut by a transversal. If one interior angle is 110°, what is the corresponding angle?
Answer & Explanation:
Corresponding angles are equal, so the corresponding angle is also 110°.
10. Three parallel lines cut two transversals at points P, Q, R and S, T, U respectively. If PQ:QR = ST:TU, what can you say about the lines?
Answer & Explanation:
If the segments are in the same ratio, the lines are parallel by the ratio theorem.