A straight line can be considered as a line passing through at least two points and remains straight all the time. In xy plane a straight line equation contains variables x and y and a constant [under certain conditions only one variable may exist].

Equation from x-axis and y-axis intercept:

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

x-axis intercepting point (a, 0) and y-axis intercepting point (0, b)

The value of a and b can be positive or negative based on the following condition:

If the line intersects at positive x-axis, then a is positive, otherwise negative. If the line intersects at positive y-axis, then b is positive, otherwise negative Following figures shows different intersecting point and corresponding straight line equation.

\; \frac{x}{-3} + \frac{y}{7} = 1 \; \frac{x}{3} + \frac{y}{7} = 1

\; \frac{x}{-3} + \frac{y}{-7} = 1 \; \frac{x}{3} + \frac{y}{-7} = 1

Equation from slope:

y = mx + c

Where m is the slope or gradient of the straight line and c is the y-axis intercept. Slope can be determined by taking two points in the straight line and calculating ratio of change in y to change in x as shown in the following figure.

If c = 0, the equation becomes y = mx This is the straight line equation passing through the origin (0,0)

Exercise:

Suppose a straight line equation be given by,

\frac{x}{10} + \frac{y}{20} = 1

What is the slope of above equation?

Solution: Multiplying both sides of equation with 20 we get,

2x + y = 20 y = -2x + 20

So slope, m= -2

Exercise:

Suppose we have a straight line equation

y = 2x + 20

What is the x-axis and y-axis intercept?

Solution: Above equation can be written as

\frac{x}{-10} + \frac{y}{20} = 1

So, the line intercepts negative x-axis at 10 and positive y-axis at 20.

Equation of line parallel to x-axis and y-axis:

Equation of a straight line parallel to x-axis is given by

y = b

Here b is the unit distance from x-axis. For example in the following figure y = 3 is parallel to x-axis with every point on this line is at 3 unit distance from x-axis.

Equation of a straight line parallel to y-axis is given by

x = a

Here a is the unit distance from y-axis. For example in the following figure x = 7 is parallel to y-axis with every point on this line is at 7 unit distance from y-axis.

Straight line equation passing through two points:

The equation of a straight line passing through the points (x₁, y₁) and (x₂, y₂) is given by

\frac{y-y_2}{y_2-y_1} = \frac{x-x_2}{x_2-x_1}

Exercise:

Derive the straight line equation passing through the points (20, 0) and (-7, 3).

Let (x₁, y₁) = (20, 0)   and      (x₂, y₂) = (-7, 3)

So, the above equation becomes

\frac{y-3}{3-0} = \frac{x-(-7)}{-7-20}

or, 3x + 21 = -27y + 81

or, 3x + 27y - 60 = 0

Straight line equation parallel to another straight line:

Straight line parallel to another straight line ax + by + c = 0 is given by ax + by + k = 0 where k is another constant which need to be determined from other conditions. All parallel straight lines have equal slope.

Exercise:

Determine the straight line equation which is parallel to 2x + 3y - 18 = 0 and passes through the point (1, -3).

Let the equation be 2x + 3y + k = 0

Since it passes through the point (1, -3), substituting this value we get,

2×1 + 3×(-3) + k = 0

or, k = 7

So the straight line is 2x + 3y + 7 = 0

Let’s  find the slope of the original line and the parallel line:

Original line: 3y = -2x + 18

or, y = -\frac{2}{3}x + 6

So slope, m_1 = -\frac{2}{3}

New Line: 3y = -2x + 7

or, y = -\frac{2}{3}x + \frac{7}{3}

So slope, m_2 = -\frac{2}{3}

Hence m_1 = m_2

Straight line equation perpendicular to another straight line:

Straight line perpendicular to another straight line ax + by + c = 0 is given by bx - ay + k = 0 where k is another constant which need to be determined from other conditions. If two lines are perpendicular to each other, then product of their slope is -1

Exercise:

Determine the straight line equation which is perpendicular to 2x + 3y - 18 = 0 and passes through the point (1, -3).

Let the equation be 3x - 2y + k = 0

Since it passes through the point (1, -3), substituting this value we get,

            3×1 –  2×(-3) + k = 0

or,      k = -9

So the straight line is 3x - 2y - 9 = 0

Let’s  find the slope of the original line and the perpendicular line

Original line: 3y = -2x + 18

or, y = -\frac{2}{3}x + 6

So slope m_1 = -\frac{2}{3}

New line: 2y = 3x - 9

or, y = \frac{3}{2}x - \frac{9}{2}

So slope m_2 = \frac{3}{2}

Hence m_1m_2 = - 1

Length of the perpendicular line drawn from (x₁, y₁) onto line ax + by + c = 0 is given by,

\frac{|ax_1 + by_1 + c|}{\sqrt(a^2 + b^2)}

Exercise:

Find the length of perpendicular from point (-3,7) to line 2x - 3y + 20 = 0

Distance = \frac{|2\cdot(-3) + (-3)\cdot7 + 20|}{\sqrt(2^2 + (-3)^2)}

= \frac{|-7|}{\sqrt{13}}

= \frac{7}{\sqrt{13}}