Notes On Slope of a Line - CBSE Class 11 Maths
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 = m2 ⇒ l1 || l2 Relationship between the slopes of perpendicular lines: Given: l1 ⊥ l2 ⇒ 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 ⇒ l1 ⊥ l2 #### 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 = m2 ⇒ l1 || l2 Relationship between the slopes of perpendicular lines: Given: l1 ⊥ l2 ⇒ 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 ⇒ l1 ⊥ l2 Previous
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