Equations of a Straight Line

Learn the different forms of the equation of a straight line and understand how to find each form step by step.

Equations of a Straight Line

An equation of a straight line represents the relationship between the coordinates $x$ and $y$ of every point lying on a straight line. Different forms of the equation are useful depending on the information given about the line.

In this topic, we will learn the following forms:

Equation of a line parallel to the Y-axis – used when the line is vertical.
Equation of a line parallel to the X-axis – used when the line is horizontal.
Slope-intercept form – used when the slope and $y$-intercept are known.
Double-intercept form – used when the $x$-intercept and $y$-intercept are known.
Perpendicular form – used when the perpendicular distance of the line from the origin and the angle made by the perpendicular with the positive $x$-axis are known.

Section Formula – Coordinates of a Point Dividing a Line Segment

Step 1 of 7

Section Formula

Let, in the figure, A(x₁, y₁) and B(x₂, y₂) are two points. P(x, y) divides the line segment AB internally in the ratio m₁:m₂. Calculate the coordinate of P(x, y).

Now, draw AD⊥OX, BE⊥OX, PF⊥OX, AM⊥PF and PN⊥BE

In the figure, AM = DF = OF − OD = x − x₁,

PN = FE = OE − OF = x₂ − x

PM = PF − MF = PF − AD = y − y₁

BN = BE − NE = BE − PF = y₂ − y

And APBP = m₁m₂

Right angled triangles ΔPMA and ΔBNP are similar.

AMPN = APBP [ ∵ The ratio of corresponding sides of similar triangle]

Or, x − x₁x₂ − x = m₁m₂

Or, m₂x − m₂x₁ = m₁x₂ − m₁x

Or, m₂x + m₁x = m₁x₂ + m₂x₁

Or, x(m₁ + m₂) = m₁x₂ + m₂x₁

Or, x = m₁x₂ + m₂ x₁m₁ + m₂

Again, PMBN = APBP [ ∵ The ratio of corresponding sides of similar triangle]

Or, y − y₁y₂ − y = m₁m₂

Or, m₂y − m₂y₁ = m₁y₂ − m₁y

Or, m₂y + m₁y = m₁y₂ + m₂y₁

Or, y(m₁ + m₂) = m₁y₂ + m₂y₁

Or, y = m₁y₂ + m₂y₁m₁ + m₂

∴ (x, y) = (m₁x₂ + m₂ x₁m₁ + m₂ , m₁y₂ + m₂ y₁m₁ + m₂)

Thus, the coordinates of the point dividing line segment AB are

P(x, y) = (m₁x₂ + m₂ x₁m₁ + m₂ , m₁y₂ + m₂ y₁m₁ + m₂)

Equation of line parallel to X- axis

Derivation Steps

In the adjoining figure, the line AB is parallel to the X- axis, and the distance from the X- axis to AB is b units.

Thus, the y- coordinate of every point lying on AB is b.

Therefore, y = b is the equation of the line AB.

If the line AB is below the X- axis and the distance from the X- axis to the line is b units, what would its equation be? Discuss this among your friends and present it in the classroom.

Equation of straight line in slope intercept form

Derivation Steps

In the adjoining figure, the straight line AB makes an angle ∠BAX = θ with the positive direction of the X- axis.

Its y- intercept is OC = (c). From any point P(x, y) lying on this line, we draw PM ⊥ OX. We also draw CN ⊥ PM. We have ∠PCN = ∠BAX = θ, OM = x and PM = y.

In right angled triangle PNC,

tanθ = PNCN

or, m = PM – NMCN

or, m = PM – OCOM

or, m = y – cx

or, y – c = mx

∴ y = mx + c

Alternative method

The coordinate of point C = (0, c)

Slope of AB = Slope of CP (why? Discuss.)

or, m = y – cx – 0

or, mx = y – c

∴ y = mx + c is the required equation.

a. If the line AB passes through origin, c = 0 and its equation be y = mx + 0

∴ y = mx

b. If the line AB is parallel to X- axis, m = tan0° = 0, and the equation of line is

y = mx + c

Or, y = 0.x + c

∴ y = c

Equation of line parallel to Y- axis

Derivation Steps

In the adjoining figure, the line AB is parallel to the Y- axis, and the distance from the Y- axis to AB is a units.

Thus, the x- coordinate of every point lying on AB is a.

Therefore, x = a is the equation of the line AB.

If the line AB is to the left side of the Y- axis and the distance from the Y- axis to the line is a units, what would its equation be? Discuss this among your friends and present it in the classroom.

Equation of straight line in double intercepts form

Derivation Steps

In the adjoining figure, the straight line AB intersects the X- axis at the point A(a, 0) and the Y- axis at the point B(0, b). Therefore, OA = a and OB = b.

Let P(x, y) be any point lying on AB.

Slope of AP = y – 0x – a = yx – a

Slope of PB = b – y0 – x = b – y0 – x

The points A(a, 0), P(x, y), and B(0, b) are collinear (lie on the same line).

Therefore, Slope of AP = Slope of PB

Or, yx – a = b – y0 – x

Or, –xy = bx – ab – xy + ay

Or, bx + ay = ab

Dividing both sides by ab:

bxab + ayab = abab

∴ xa + yb = 1

This is the equation of the line AB in the double intercepts form.

Equation of straight line in perpendicular form

Derivation Steps

In the given figure, the straight line AB intersects the X- axis at the point A(a, 0) and the Y- axis at the point B(0, b). Therefore, OA = a and OB = b.

A perpendicular line OD ⊥ AB is drawn, where OD = p and ∠DOA = α.

In the right-angled triangle ΔODB, ∠BOD = 90° – α, ∠OBD = 180° – (90° – α) – 90° = α.

sinα = ODOB = pb

or, b = psinα

In the right-angled triangle ΔODA,

cosα = ODOA = pa

or, a = pcosα

Now, the equation of line AB from the intercepts form

xa + yb = 1

or, xpcosα + ypsinα = 1

or, xcosαp + ysinαp = 1

Thus, xcosα + ysinα = p is the required equation of line AB.

Example 1 of 2

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