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Slope of a Line

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Slope of a Line - Lesson Summary

In a coordinate plane with two points A (x1, y1) and B (x2, y2),

Distance formula: AB = √(x2 - x1)2 + (y2 - y1)2

Ratio formula: Coordinates of a point C dividing line segment AB internally in the ratio m:n = [mx2 + nx1/m + n , my2 + ny1/m + n].

If m = n, coordinates of C = [x1 + x2/2, y1 + y2/2].

Area of triangle: Area of ∆ABC = ½ ∣x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)∣

If area of ∆ABC = 0 ⇒ A, B and C are collinear points.


A line is said to be inclined when it makes an angle with the horizontal. A line intersecting the X-axis forms supplementary angles.

The angle made by a straight line in the anti-clockwise direction with the X-axis is called inclination.

The anti-clockwise direction is also called the positive direction. The X- axis and the lines parallel to it are called horizontal lines

If line AB lies along the X-axis or is parallel to the X-axis, then its inclination is zero. The Y-axis and the lines parallel to it are called vertical lines.

If line AB is parallel to the Y-axis or perpendicular to the X-axis, then its inclination is 90°. The inclination of a line can have a value anywhere from zero to 180°.

Slope of line AB (m) = tan θ

If θ = 0o

m = tan 0o = 0

If θ = 90o

m = tan 90o ⇒ Not defined.

Slope of a line passing through two given points:

Let P (x1, y1) and Q (x2, y2) be two points on a non-vertical line l, whose inclination is θ.

Since the line is not a vertical line, x1 ≠ x2.

Case I: 0o ≤ θ < 90o

Draw perpendiculars 'QR' and 'PT' from 'Q' and 'P' on to the X-axis.

Draw perpendicular 'PM' from 'P' on to 'QR'.

Since QR ⊥ PM and QR ⊥ X-axis,

PM || X-axis

⇒ ∠QSX = ∠QPM = θ (Corresponding angles)

In ∆PMQ:

tan θ = QM/PM ………………… (1)

We have

QM = QR - MR

QM = y2 - y1 ………………… (2)

PM = TR

PM = OR - OT

PM = x2 - x1 …………………… (3)

Substituting the values of QM and PM from (2) and (3) in (1):

Tan θ = (y2 - y1)/(x2 - x1).

Case II: 90o < θ ≤ 180o

Draw perpendiculars 'QR' and 'PT' onto X-axis.

Draw perpendicular 'PM' from 'P' to 'QR' and extend 'MP' to 'S'.

Since QR ⊥ PM and QR ⊥ X-axis,

PM || X-axis

⇒ ∠QUX = ∠QPS = θ (Corresponding angles)

∠QPS + ∠QPM = 180o (Supplementary angles)

∴ ∠QPM = 180o – θ

⇒ θ = 180 – ∠QPM

Thus, m = tan θ= tan(180 - ∠QPM)

m = – tan ∠QPM ….(1)

In ∆PMQ:

tan ∠QPM = QM/PM = QM/RT

⇒ tan ∠QPM = (y2 - y1)/(x2 - x1) ….(2)

From equations 1 and 2:

m = -(y2 - y1)/(x2 - x1)

= (y2 - y1)/(x2 - x1)

m = (y2 - y1)/(x2 - x1)

Conditions for parallelism and perpendicularity

Consider two non-vertical parallel lines l1 and l2 having inclination a and b, respectively.

If line l1 and l2 are parallel, their inclination must be the same.

⇒ α = β

⇒ tan α = tan β

tan α = m1 and tan β = m2

m1 = m2

Thus, we can say that if two non-vertical lines are parallel, their slopes are equal.

The converse is also true. That is, if the slopes of two non-vertical lines are equal, the two lines are parallel.

If m1 = ml1 || l2

Relationship between the slopes of perpendicular lines:

Given: l1l2

⇒ β = α + 90o

⇒ tan β = tan (α + 90o)

⇒ tan β = - cot α

⇒ tan β = - 1/tan α …..Equation (1)

tan α = m1 and tan β = m2

m2 = -1/m1

m1 m2 = -1

Given: m1 m2 = -1

l1l2

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