Introduction to Coordinate Geometry
Coordinate geometry (also called analytic geometry) was developed by Rene Descartes (1596-1650). It establishes a relationship between algebra and geometry using coordinate axes.
Slope of a Line
The slope (or gradient) of a line is a measure of its steepness and direction.
If a line passes through points P(x_1, y_1) and Q(x_2, y_2):
m = (y_2 - y_1)/(x_2 - x_1), where x_1 != x_2
Types of Slope
- Positive slope: line rises from left to right (
m > 0) - Negative slope: line falls from left to right (
m < 0) - Zero slope: horizontal line (
m = 0) - Undefined slope: vertical line (denominator zero)
Slope Intercept Form
If theta is the angle a line makes with positive x-axis, m = tan theta.
Angle Between Two Lines
If m_1 and m_2 are slopes of two lines, the acute angle theta between them:
tan theta = |(m_2 - m_1)/(1 + m_1 m_2)|
Parallel Lines
Lines are parallel if m_1 = m_2.
Perpendicular Lines
Lines are perpendicular if m_1 * m_2 = -1 (product of slopes is -1).
Various Forms of Equations
Point-Slope Form
Line through (x_1, y_1) with slope m:
y - y_1 = m(x - x_1)
Two-Point Form
Line through (x_1, y_1) and (x_2, y_2):
(y - y_1)/(y_2 - y_1) = (x - x_1)/(x_2 - x_1) or y - y_1 = ((y_2 - y_1)/(x_2 - x_1))(x - x_1)
Slope-Intercept Form
Line with slope m and y-intercept c:
y = mx + c
Intercept Form
Line with x-intercept a and y-intercept b:
x/a + y/b = 1
Normal Form
If p is perpendicular distance from origin and omega is angle of the perpendicular:
x cos omega + y sin omega = p
General Form
Ax + By + C = 0, where A, B are not both zero.
Distance of a Point from a Line
Distance from point (x_1, y_1) to line Ax + By + C = 0:
d = |Ax_1 + By_1 + C|/sqrt(A^2 + B^2)
Distance Between Parallel Lines
For lines Ax + By + C_1 = 0 and Ax + By + C_2 = 0:
d = |C_1 - C_2|/sqrt(A^2 + B^2)
Concurrent Lines
Three lines are concurrent if they pass through a common point.
Condition for concurrency: The intersection point of any two lines must satisfy the third equation.
Alternatively, the determinant condition:
|A_1 B_1 C_1|
|A_2 B_2 C_2| = 0
|A_3 B_3 C_3|
Worked Examples
Example 1: Find the slope of the line passing through (2, 3) and (5, 7).
Solution: m = (7-3)/(5-2) = 4/3.
Example 2: Find the equation of the line with slope 2 passing through (3, -1) in point-slope form.
Solution: y - (-1) = 2(x - 3) => y + 1 = 2x - 6 => y = 2x - 7.
Example 3: Find the equation of a line passing through (2, 3) and (4, 7).
Solution: m = (7-3)/(4-2) = 2. Using point-slope: y - 3 = 2(x - 2) => y = 2x - 1.
Example 4: Find the distance of point P(3, 4) from the line 3x + 4y - 10 = 0.
Solution: d = |3(3) + 4(4) - 10|/sqrt(9+16) = |9+16-10|/5 = 15/5 = 3 units.
Example 5: Find the lines parallel and perpendicular to y = 2x + 3 passing through (1, 2).
Solution:
Parallel (same slope m=2): y - 2 = 2(x-1) => y = 2x.
Perpendicular (slope m = -1/2): y - 2 = (-1/2)(x-1) => 2y - 4 = -x + 1 => x + 2y = 5.
Common Mistakes
- Slope of vertical line: It is undefined, not 0.
- Perpendicular condition:
m_1 * m_2 = -1, notm_1 = -m_2. - Intercept form sign:
x/a + y/b = 1. Intercepts are signed quantities. - Distance formula absolute value: Distance is always positive, use modulus.
Transformation Between Forms
- General to slope-intercept:
y = (-A/B)x + (-C/B), som = -A/B,c = -C/B. - General to intercept:
x/(-C/A) + y/(-C/B) = 1, soa = -C/A,b = -C/B.
ISC Exam Focus
- Theory (70%): Slope, angle between lines, distance formula derivations, standard forms.
- Application (30%): Equation finding, concurrency problems, distance calculations.
- Typical ISC questions: "Find the equation of line passing through ... and perpendicular to ...".
- 4-6 mark questions often combine multiple concepts.
Self-Test Questions
Q1: Find the slope of the line making an angle of 60 degrees with the positive x-axis.
Answer: m = tan 60 = sqrt(3).
Q2: Find the equation of a line with x-intercept 4 and y-intercept 5.
Answer: x/4 + y/5 = 1 => 5x + 4y = 20.
Q3: Find the angle between lines y = 3x + 2 and y = -x + 5.
Answer: m_1 = 3, m_2 = -1. tan theta = |(-1-3)/(1+3(-1))| = |(-4)/(-2)| = 2. theta = tan^(-1)(2).
Q4: Show that lines 2x + 3y - 5 = 0, 4x + 6y + 7 = 0 are parallel.
Answer: m_1 = -2/3, m_2 = -4/6 = -2/3. Since m_1 = m_2, lines are parallel. Distance = |(-5) - (7/2)|/sqrt(4+9) = |-17/2|/sqrt(13) = 17/(2sqrt(13)).
Q5: Find the foot of perpendicular from (1, 2) to line 2x + y = 5.
Answer: Let foot be (h, k). Then (h-1)/2 = (k-2)/1 = -(2(1)+2-5)/(4+1) = -(-1)/5 = 1/5. So h = 1 + 2/5 = 7/5, k = 2 + 1/5 = 11/5. Foot = (7/5, 11/5).
Q6: Find k if lines 2x + 3y + 4 = 0, 3x + 4y - 5 = 0, and 4x + 5y + k = 0 are concurrent.
Answer: Intersection of first two: solving gives x = -31, y = 22. Substituting in third: 4(-31) + 5(22) + k = 0 => -124 + 110 + k = 0 => k = 14.
