SMA 103 • UNIT 1 • Mathematics

The Straight Line

Topics covered: Cartesian Coordinate System and Distance Formula, Internal and External Division of Line Segments, Gradient, Slope-Intercept, and Normal Forms, Parallel and Perpendicular Lines, Angle Between Two Intersecting Lines, Perpendicular Distance from a Point to a Line

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1. Core Concept Overview & Intuition

The straight line is the simplest and most fundamental curve in analytical geometry. Its study begins with the Cartesian coordinate system, which provides a systematic way to describe the position of any point in a plane by an ordered pair of numbers. Once points can be located precisely, one can measure the distance between them, locate points that divide a segment in a given ratio, describe the steepness and direction of a line, and write equations that capture an entire infinite set of collinear points. These ideas form the foundation for all later work in analytical geometry, because every more complicated curve is studied by relating it to lines, distances, and angles.

The central intuition is that a straight line is completely determined by two pieces of information: a point through which it passes and its direction, or equivalently two distinct points through which it passes. Direction is quantified by the slope, which measures the change in the vertical coordinate per unit change in the horizontal coordinate. From the slope and a single point, an equation for the line can be written. Different algebraic arrangements of that equation—point-slope form, slope-intercept form, general form, and normal form—emphasise different geometric features such as the intercepts or the perpendicular distance from the origin.

Two lines in the same plane either intersect at a single point or are parallel. Their slopes encode this relationship: equal slopes mean parallel lines, while slopes whose product is negative one mean perpendicular lines. When two lines do intersect, the angle between them can be computed from their slopes using the tangent function. Finally, the perpendicular distance from a point to a line measures the shortest separation between a point and a line, a quantity that appears repeatedly in geometry and in applications.

Throughout this topic, every formula is derived from the coordinate system and the right-angle triangle. The distance formula is an application of Pythagoras' theorem, the slope is a ratio of coordinate differences, and the angle between two lines is obtained from the tangent of a difference of inclinations. Mastery of these derivations, rather than mere memorisation of formulas, is the goal.

1.2.1.4 Parallel and Perpendicular Lines If two lines and are parallel, then their slopes are equal;

Figure 1.6 Given that the slope of is and the slope of is , then the product of their slopes is -1; 1.2.1.5 The Angle between Two Lines

2. Mathematical Formalization & Core Definitions

2.1 The Cartesian Coordinate System

A point on the plane is described by its abscissa and its ordinate as $P(x,y)$, for any point $P$. The symbol $(x,y)$ represents the coordinates of the point, where $x$ is the abscissa and $y$ the ordinate; this ordered pair is also referred to as the Cartesian coordinates of $P$. The two axes partition the plane into four quadrants. In the first quadrant both $x$ and $y$ are positive. In the second quadrant $x$ is negative while $y$ is positive. In the third quadrant both $x$ and $y$ are negative. In the fourth quadrant $x$ is positive and $y$ is negative. The coordinates of the origin are $(0,0)$.

point on the plane can described by its abscissa and its ordinate as P(x, y), for any point P.

2.2 The Distance Between Two Points

The distance between two points $P(x_1,y_1)$ and $Q(x_2,y_2)$ on a straight line is given by

$$
PQ=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}.
$$

This result follows from constructing a right-angled triangle whose horizontal leg has length $|x_2-x_1|$ and whose vertical leg has length $|y_2-y_1|$, and then applying Pythagoras' theorem.

Example 1.1. Find the distance between the points $P(2,3)$ and $Q(5,7)$.

Applying the distance formula,

$$
PQ=\sqrt{(5-2)^2+(7-3)^2}=\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5.
$$

2.3 The Mid-point of a Line

The coordinates of the mid-point of a line between two end points $P(x_1,y_1)$ and $Q(x_2,y_2)$ are

$$
\left(\frac{x_1+x_2}{2},\ \frac{y_1+y_2}{2}\right).
$$

The mid-point of the straight line joining $P$ and $Q$ is therefore obtained by averaging the abscissae and averaging the ordinates separately.

Figure 1.2 The distance between the points is; 1.2.1.2 The Mid-point of a Line The co-ordinates of the mid-point of a line between two end points and is; .

2.4 Internal and External Division of a Line Segment

The mid-point is the special case of dividing a segment in the ratio $1:1$. More generally, a point $R$ that divides the segment $PQ$ internally in the ratio $m:n$ has coordinates

$$
R=\left(\frac{mx_2+nx_1}{m+n},\ \frac{my_2+ny_1}{m+n}\right).
$$

When the point divides the segment externally in the ratio $m:n$, the point lies on the line through $P$ and $Q$ but outside the segment, and its coordinates are

$$
R=\left(\frac{mx_2-nx_1}{m-n},\ \frac{my_2-ny_1}{m-n}\right).
$$

The internal division formula places the point between the two end points, whereas the external division formula places it beyond one of the end points. Setting $m=n=1$ in the internal formula recovers the mid-point formula.

2.5 Inclination and Slope of a Line

The inclination of a line is the angle $\theta$ that the line makes with the positive direction of the $x$-axis, measured anticlockwise. The slope (or gradient) $m$ of the line is the tangent of the inclination:

$$
m=\tan\theta.
$$

For a line passing through two distinct points $P(x_1,y_1)$ and $Q(x_2,y_2)$, the slope is

$$
m=\frac{y_2-y_1}{x_2-x_1}.
$$

Considering the right-angled triangle formed by the horizontal and vertical differences between the two points, the tangent of the inclination equals the ratio of the vertical difference to the horizontal difference.

Example 1.2. Find the slope and the angle of inclination of the line passing through the points $P(1,2)$ and $Q(4,8)$.

The slope is

$$
m=\frac{8-2}{4-1}=\frac{6}{3}=2.
$$

Since $m=\tan\theta=2$, the angle of inclination is $\theta=\tan^{-1}(2)$, which is approximately $63.4^\circ$.

Considering the right angle triangle, then we have; Example 1.2 Find the slope and the angle of inclination of the line passing through points and .

2.6 Parallel and Perpendicular Lines

If two lines $L_1$ and $L_2$ are parallel, then their slopes are equal:

$$
m_1=m_2.
$$

If the two lines are perpendicular, then the product of their slopes is $-1$:

$$
m_1m_2=-1.
$$

Equivalently, the slope of a line perpendicular to a line of slope $m$ is $-1/m$.

1.2.1.4 Parallel and Perpendicular Lines If two lines and are parallel, then their slopes are equal;

2.7 The Equation of a Straight Line

Suppose that $P(x_1,y_1)$ and $Q(x_2,y_2)$ are points on a line. The slope of the line is

$$
m=\frac{y_2-y_1}{x_2-x_1}.
$$

Taking the general point $(x,y)$ on the line, the point-slope form of the equation of the line is

$$
y-y_1=m(x-x_1).
$$

Rearranging gives the slope-intercept form

$$
y=mx+c,
$$

where $c$ is the $y$-intercept, that is, the ordinate of the point where the line crosses the $y$-axis.

The general form of the equation of a straight line is

$$
ax+by+c=0,
$$

where $a$, $b$, and $c$ are constants and $a$ and $b$ are not both zero. The normal form of the equation of a straight line expresses the line in terms of the perpendicular distance $p$ from the origin and the angle $\alpha$ that this perpendicular makes with the positive $x$-axis:

$$
x\cos\alpha+y\sin\alpha=p.
$$

The point-slope, slope-intercept, general, and normal forms are different algebraic representations of the same geometric object; each is convenient for a different type of problem.

Figure 1.8 Then the slope of the line is; Taking the general point , we get the equation of the line to be; Where is the y-intercept Example 1.4

2.8 The Angle Between Two Lines

Let two lines $L_1$ and $L_2$ meet at an angle $\theta$. Let the slope of $L_1$ be $m_1$ and the slope of $L_2$ be $m_2$. The angle between the two lines satisfies

$$
\tan\theta=\frac{m_2-m_1}{1+m_1m_2}.
$$

The angle obtained is the acute angle between the lines. If $1+m_1m_2=0$, then $\tan\theta$ is undefined, which corresponds to the lines being perpendicular.

Example 1.3. The angle between two lines is $45^\circ$. If the slope of $L_1$ is $2$, determine the slope of $L_2$.

Using the angle formula with $\theta=45^\circ$, so that $\tan\theta=1$,

$$
1=\frac{m_2-2}{1+2m_2}.
$$

Cross-multiplying gives $1+2m_2=m_2-2$, hence $m_2=-3$. The slope of $L_2$ is therefore $-3$.

2.9 The Perpendicular Distance from a Point to a Line

The perpendicular distance $d$ from a point $P(x_1,y_1)$ to the line $ax+by+c=0$ is the length of the perpendicular segment drawn from the point to the line. This distance is given by

$$
d=\frac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}.
$$

The numerator measures the extent to which the coordinates of the point fail to satisfy the equation of the line, and the denominator normalises this quantity so that the result is a genuine length. The absolute value ensures that the distance is non-negative regardless of which side of the line the point lies on.

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Question 1 (6 marks)

Find the equation of the straight line:

(i) passing through the point $(2,-1)$ and perpendicular to the line $4x-3y+7=0$;
(ii) passing through the point $(3,2)$ and parallel to the line $2x+5y-1=0$;
(iii) making intercepts $4$ on the $x$-axis and $-2$ on the $y$-axis.

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Question 2 (20 marks)

a) Solve $\tan^2 x + 3 \sec x = 0$ \newline
$0^\circ \le x < 360^\circ$ [4marks]

b) Show that \newline
$2 \sin 45^\circ \cos 15^\circ = \frac{\sqrt{5} - 1}{2}$ [4marks]

c) Find the value of $\tan 15^\circ$ without using a table or a calculator. [4marks]

d) Verify the identities

i) \quad $1 - \frac{\cos^2 \theta}{1 + \sin \theta} = \sin \theta$ [4marks]

ii) \quad $\cot \theta + \frac{\sin \theta}{1 + \cos \theta} = \csc \theta$ [4marks]

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