Chapter 9: Straight Lines

9.1 Introduction

Coordinate geometry involves the study of geometry using a coordinate system. This bridges the gap between algebra and geometry.

Important formulas from earlier classes:

• Distance Formula: The distance between points \( P(x_1, y_1) \) and \( Q(x_2, y_2) \) is \( PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \).
• Section Formula: The coordinates of a point dividing the line segment joining \( (x_1, y_1) \) and \( (x_2, y_2) \) internally in the ratio \( m:n \) are \( \left(\frac{mx_2 + nx_1}{m + n}, \frac{my_2 + ny_1}{m + n}\right) \).
• Area of Triangle: Area = \( \frac{1}{2}|x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| \).

9.2 Slope of a Line

If \( \theta \) is the angle of inclination of a line \( l \), then \( \tan \theta \) is called the slope or gradient of the line \( l \). Slope is usually denoted by \( m \). \[ m = \tan \theta, ; \text{where } \theta \neq 90^\circ \] The slope of a line passing through points \( (x_1, y_1) \) and \( (x_2, y_2) \) is: \[ m = \frac{y_2 - y_1}{x_2 - x_1}, ; \text{where } x_1 \neq x_2 \]

Angle between Two Lines

If \( m_1 \) and \( m_2 \) are slopes of two non-vertical lines, the acute angle \( \theta \) between them is: \[ \tan \theta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right| \]

• Lines are parallel if \( m_1 = m_2 \).
• Lines are perpendicular if \( m_1 m_2 = -1 \).

9.3 Various Forms of Equations of a Line

1. Horizontal and vertical lines: Equation of x-axis is \( y = 0 \). Equation of y-axis is \( x = 0 \).
2. Point-slope form: Equation of a line passing through \( (x_0, y_0) \) with slope \( m \) is: \[ y - y_0 = m(x - x_0) \]
3. Two-point form: Line passing through \( (x_1, y_1) \) and \( (x_2, y_2) \) is: \[ y - y_1 = \frac{y_2 - y_1}{x_2 - x_1} (x - x_1) \]
4. Slope-intercept form: Line with slope \( m \) and y-intercept \( c \) is \( y = mx + c \).
5. Intercept form: Line making intercepts \( a \) and \( b \) on x and y axes is: \[ \frac{x}{a} + \frac{y}{b} = 1 \]
6. Normal form: Line uniquely determined by length of perpendicular \( p \) from origin and angle \( \omega \) this perpendicular makes with positive x-axis: \[ x \cos \omega + y \sin \omega = p \]

9.4 General Equation of a Line

Any equation of the form \( Ax + By + C = 0 \), where \( A \) and \( B \) are not simultaneously zero, is called the general linear equation. It always represents a straight line.

• Slope \( m = -\frac{A}{B} \)
• x-intercept \( = -\frac{C}{A} \)
• y-intercept \( = -\frac{C}{B} \)

Distance of a Point From a Line

The perpendicular distance \( d \) from a point \( (x_1, y_1) \) to a line \( Ax + By + C = 0 \) is: \[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \] Distance between two parallel lines \( Ax + By + C_1 = 0 \) and \( Ax + By + C_2 = 0 \): \[ d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}} \]

Competency-Based Questions

Question 1:

A line passes through the point \( P(3,4) \) and cuts off intercepts on the coordinate axes such that their sum is 14. Determine the possible equation(s) of this line.

Answer 1:

Let the intercepts on the x and y axes be \( a \) and \( b \) respectively.

The equation of the line in intercept form is:

\[ \frac{x}{a} + \frac{y}{b} = 1 \]

We are given that \( a + b = 14 \Rightarrow b = 14 - a \).

Also, the line passes through \( (3, 4) \), so we substitute \( x=3 \) and \( y=4 \):

\[ \frac{3}{a} + \frac{4}{14 - a} = 1 \]

Take the LCM:

\[ \frac{3(14 - a) + 4a}{a(14 - a)} = 1 \]

\[ 42 - 3a + 4a = 14a - a^2 \]

\[ 42 + a = 14a - a^2 \]

Rearranging to form a quadratic equation:

\[ a^2 - 13a + 42 = 0 \]

Factorizing the quadratic equation:

\[ (a - 6)(a - 7) = 0 \]

So, \( a = 6 \) or \( a = 7 \).

Case 1: If \( a = 6 \), then \( b = 14 - 6 = 8 \).

The equation is \( \frac{x}{6} + \frac{y}{8} = 1 \Rightarrow 4x + 3y = 24 \).

Case 2: If \( a = 7 \), then \( b = 14 - 7 = 7 \).

The equation is \( \frac{x}{7} + \frac{y}{7} = 1 \Rightarrow x + y = 7 \).

Conclusion: The two possible equations for the line are \( 4x + 3y = 24 \) and \( x + y = 7 \).

Question 2:

Find the angle between the lines \( y - \sqrt{3}x - 5 = 0 \) and \( \sqrt{3}y - x + 6 = 0 \).

Answer 2:

First, rewrite both equations in slope-intercept form (\( y = mx + c \)) to find their slopes.

Line 1: \( y = \sqrt{3}x + 5 \)

Slope \( m_1 = \sqrt{3} \)

Line 2: \( \sqrt{3}y = x - 6 \Rightarrow y = \frac{1}{\sqrt{3}}x - \frac{6}{\sqrt{3}} \)

Slope \( m_2 = \frac{1}{\sqrt{3}} \)

The acute angle \( \theta \) between the lines is given by:

\[ \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \]

\[ \tan \theta = \left| \frac{\sqrt{3} - \frac{1}{\sqrt{3}}}{1 + (\sqrt{3})\left(\frac{1}{\sqrt{3}}\right)} \right| \]

Simplify numerator:

\[ \sqrt{3} - \frac{1}{\sqrt{3}} = \frac{3 - 1}{\sqrt{3}} = \frac{2}{\sqrt{3}} \]

Simplify denominator:

\[ 1 + 1 = 2 \]

Thus,

\[ \tan \theta = \left| \frac{ \frac{2}{\sqrt{3}} }{2} \right| = \frac{1}{\sqrt{3}} \]

The angle whose tangent is \( \frac{1}{\sqrt{3}} \) is \( 30^\circ \) or \( \frac{\pi}{6} \) radians.

Conclusion: The angle between the two lines is \( 30^\circ \).